«i 



i! 



ii 



PRACTICAL MECHANICS, 



THE 



APPRENTICE 



R 



FIRST BOOK 



FOR 



MECHANICS, MACHINISTS, 



AND 



ENGINEERS. 



BY 

OLIVER BYRNE, 

MATHEMATICIAN, 
CIYIL, MILITARY, AND MECHANICAL EXGiXEER. 

Author of " The Practical Model Calculator ;^^ Compiler and Editor of the " Dictionary 

of Machines, Mechanics, Engine-work, and Engineering/;'''' '^ The Pocket Companion, 

' for Machinists, Mechanics, and Engineers ;^^ ''The Practical Cotton- Spinner ;^^ 

'- The Practical Metal-worker^ s Assistant ;^' author and inventor of the 

" Calculus offl'hi-m,'" a new Scienc-a, a sidjstitute for the Differential 

and Integral Calculus; '■'■ Th< Doctrine of Proportion ;''^ 

" The Elements of Euclid, hy Colours, etc , etc., 



N E W Y RK: 
PUBLISHED BY PHILIP J. COZANS, 

No. 107 NASSAU STREET CORNER OF ANN. 
1860. 



? S^ 



^..^i^- ^//J^^ 



(i-- 



A 



Entered according to Act of Congress, in the year 1859. 

By PHILIP J. COZANS, 

In the Clerk's Office of the District Court of the United States, for the Southern 

District of New York. 



STEREOri'PED BY 

VINCENT L. DILL, 
128 Fulton Street, (Sun Building) . 



iH-W 



PREFACE. 



There being no Elementary treatise on Practical Me- 
chanics in the English language, the present work is de- 
signed to supply the deficiency. It may be asked how can 
this be, while there exist so many elementary treatises on 
the theory and application of Mechanics, or why not ft-ans- 
late, or copy the great French works, as the German and 
other nations do. To understand the elementary treatises 
on the theory and application of Mechanics employed as 
class books in our schools and colleges, requires consider- 
able mathematical skill, and after all, the elementary know 
ledge acquired is of little value to the practical man. On 
the general theory of Mechanics, there are also many popu- 
lar works clear of mathematical formula and serviceable to 
the general reader, but of no practical value. French ele- 
mentary writers on practical Mechanics, take for granted 
that they are writing for those who understand the theory 
of Mechanics, and when these works are copied or trans- 
lated, they are far from being elementary. 

As practical mechanics is based upon what is termed the 
principle of work, to be able to distinguish with ease, units 
of work from other units, would greatly facilitate the acqui- 
sition of this branch of knowledge ; I have, therefore, in- 
troduced a new and simple notation to effect this object, 
which FreixCh writers and their copyists have neglected. 

The unit of work is equal to the labor required to raise 



VI PREFACE. 

a pound weight through*tlie space of one foot, against the 
direct action of the force of gravity. If a man takes a 
pound weight in his hand, and raise it one foot in the di- 
rection of a plumb-line, he will perform a unit of work. 

1 lb. raised 24 feet high = 24 units of work 

2 lbs. raised 12 feet high = 24 units of work . 

3 lbs. raised 8 feet high = 24 units of work 

4 lbs. raised 6 feet high = 24 units of work 
6 lbs. raised 4 feet high = 24 units of work 
8 lbs. raised 3 feet high = 24 units of work 

12 lbs. raised 2 feet high = 24 units of work 
24 lbs. raised 1 foot high = 24 units of work. 
To distinguish units of work from other units, I simply 
write or set down the numbers that represent them, inclin- 
ing to the left, instead of the ordinary way, thus : 
^4: represents 24 units of work ; 
8 X 3 = *i4 ; 

that is the same as saying, 8 lbs. raised 3 feet high is equal 
to 24 units of work. 

^4:' ; represents 24 units of work done in a minute. 

^4'^ ; represents 24 units of work done in a second. 
33000 =^11^; 
represents 33,000 units of work done in a minute, equal one 
horse power. 

I have been constantly asked the question, by those 
whose design it was to become machinists and engineers, 
" What is the best elementary work on Practical Me- 
chanics ?'' ; instead of answering, *' there is no such work 
in the English language,^' I can now say, buy '' The Ap- 
PRENTICE,'' the title given to this work, to intimate that it 
is designed for persons who wish to acquire a knowledge 

of Practical Mechanics. 

OLIVER BYRNE. 



PEACTICAL MECHANICS 



CHAPTER I. 



On the unit of work with, and without reference to tae unit 

of time. 

Work is measured by a unit, like length, weight, time, 
&c. To raise one pound a foot high against the earth^s 
gravity is a unit of work ; it is clear, then, to raise 5 lbs. 
a foot high is to perform 5 units of work, and to raise 5 
lbs. four feet high equal 20 units of work. Since resistance 
and pressure of every kind may be expressed in pounds, it 
follows that the unit here described may be made to mea- 
sure every kind of work. It will be presently shown that 
a unit of work is performed whenever one pound pressure 
is exerted through a space of one foot, no matter in what 
direction the space may lie. 

In developing this subject it is necessary to distinguish 
units of work from other units ; this I propose to do by 
writing or setting down the numbers that represent them 
inclining to the left instead of in tlic ordinary way, thus, 

*i5 represents 25 units of work. 
^5' will represent 25 units of work done in a minute. 
13" will represent 73 units of work done in a second. 



8 PRAjCTICAL MECHANICS. 

Sometimes the minute is taken for the unit of time, and 
33 minutes is marked 33^ ; in other cases, the second is 
taken as the unit of time, and 45 seconds is marked 45". 

45" represents 45 seconds ; 

but ^5" units of work done in a second of time. 

a' represents a minutes, but a' represents a units of 
work done in one minute. 

EXAMPLES. 

Question I. Required the units of work expended in rais- 
ing a weight of 50 lbs. to the height of 31 feet. 
Units of work in raising 1 lb. 81 feet = SI. 
.-. 50 X 31=155t>. 

the required units of work. 

Ques. 2. The ram of a pile-driving engine weighs half a 
ton, and has a fall of 17 feet, how many units of work are 
performed in raising this ram ? 

Half a ton = 1120 lbs. 
Units of work in raising 1 lb. IT feet = n. 
.-. 1120X n = 1904c0, 

the units of work required to raise 1120 lbs. to a height of It 
feet. 

Ques. 3. How many units of work are required to raise 
7 cwt. of coal from a pit whose depth = 13 fathoms ? 
7 X 112 = 784 lbs. 
13 X 6 = 78 feet, in 18 fathoms. 

Hence the work consists in raising 784 lbs. to the height of 78 
feet. 

••. 784 X 78 = 6115^. 

Ques. 4. If the weight of a man be 183 lbs., and if he as- 
cends a perpendicular height of 20 feet, he does work in 
raising himself, what are the number of units ? 

In this operation the man raises the weight of his own body, 

.•. 183 X 20 = 3«6O. 

amount of work. 



PRACTICAL MECHANICS. 9 

If this man were to descend in a bucket, it is clear, he would 
perform the same work upon a counterpoise weight when he has 
descended 20 feet. 

Ques. 5. How many units of work will be required to 
pump 8000 cubic feet of water from a mine whose depth = 
500 fathoms ? 

A cubic foot of water weighs 62.5 lbs. 

8000 X 62.5 = 500,000 lbs. 
500 fathoms = 3000 feet. 

Consequently, the work is to raise 500,000 lbs. a perpendicular 
height of 3000 feet. 

Work = 500,000 X 3000 = 1500000000. 

Ques. 6. A horse draws 150 lbs. out of a well, by means 
of a rope going over a fixed pulley, moving at a rate 2| 
miles an hour, how many units of work does this horse per- 
form a minute, the friction being neglected ? 



2i X 5280 



= 220 feet passed over per minute. 



60 

•. Work per minute = 150 X 220 = 33000'. 
Work per second := 500'^ 



THE UNIT OF WORK REFERRED TO A UNII OF TIME. 

A unit of work, or 1, represents 1 lb. raised 1 foot. 

A unit of work in a minute, or 1', represents 1 lb. raised 
a foot high in a minute. 

A unit of work in a second, or 1'', represents 1 lb. 
raised a foot high in a second. 

It has been assumed that a horse is capable of raising 
33000 Ids. a foot high in a minute, or to perform 33000 
units of work in a minute, 

Hence a horse power = 33000' = 1 ^• 

Whether this power is greater or less than the power of 
a horse it matters little, w^hile it is a power so well defined. 

To raise 10,000 lbs. a foot high in a minute, would be a 
more convenient unit to measure bv. 



10 PRACTICAL MECHANICS. 



EXAMPLES. 



Question 1. How many horse power would it take to 
raise 3 cwt. of coal per minute from a pit whose depth = 
110 fathoms ? 

Depth =: 1 1 X 6 = 660 feet. 

3 cwt. = 112 X 3 = 336 lbs. 

660 X336=:^^1'560^ 

Since a horse power := 3300©' 

^^^ ^^ -- & ^. il!i, the required power. 

Ques. 2. How many horse power is required to raise 
2200 cubic feet of water an hour from a mine whose depth 
= 63 fathoms ? 

13^500 — 2200 X 62.5 = weight of water in lbs. 

63 X 6 = 378 feet, depth of mine. 

137500 X 378 



60 

866250 



84>6*i50'. 



= ^6i-^- 



33000 

The proposed notation, must be borne in mind, 866^50' sig- 
nifies 866250 lbs., raised one foot high in a minute, and *i6i ^^-y 
represents 26^ horse power, or 26^ times 33000 lbs. raised 1 foot 
high in a minute. 

Ques. 3. A winding engine is moved by 4 horses, what 
weiglit of coal will be raised per hour from a pit whose 
depth = 200 feet ? 

Work of the 4 horses in an hour, 

= 4 X 33000 X 60 = ^»aoooo. 

Work in raising 1 lb. of coal 200 feet = ^0«. 

Since it takes 200 units of work to raise a pound of coals, 

7920000 ^^_^ „ 
^^^ --=: 39600 lbs. 

Consequently four horses, will raise 39600 lbs. of coal or of 
any thing else, a height of 200 feet in an hour. 



PRACTICAL MECHANICS. 11 

Ques. 4. In what time will an engine of 10 horse power, 
raise 5 tons of material from the depth of 132 feet ? 

5 tons =11200 lbs. 

Work of the engine per minute, 

=: 330000^ = 33000 X 10. 

Work of raising 5 tons, 

= 11200 X 132 = 14:T[8>400, 

Because the engine performs 330000 units of work a minute, 

1478400 



330000 



4.48 minutes. 



Ques, 5. How many cubic feet of water will an engine of 
36 horse power (36 ^^•) raise in an hour from a mine whose 
• depth is 40 fathoms ? 

40 X 6 = 240 feet. 
Work in an hour, 

= 36 X 33000 X 60 = 11^80000. 
Work to raise a cubic foot of water 240 feet, 
= 240 X 62.5 = 15000. 

^l^8oooo 



15000 



= 4752 cubic feet. 



Ques, 6. From what depth will an engine of 22 horse 
]:)ower, raise 13 tons of coal in an hour ? 

Work done by the engine in an hour, 

22 X 33000 = ^^6000. 

13 tons = 29120 lbs. 



29120 



r-: 24.9 feet. 



Ques. 7. An engine is observed to raise 7 tons of mate- 
rial per hour, from a mine whose depth is 85 fathoms ; re- 
quired the horse power of the engine, supposing | of its 
work to be lost in transmission ? 

N „ , . 2240X^X85X6 ..bo^^^ 

Work per mm. z^^ ^ "^ 133^80. 



12 PRACTICAL MECHANICS. 

Since f of the work of fhe engine only go to raise the material, 
.*. I of 33000' = ^^500 , 

the units of useful work of one horse power a minute. 



*n5o«' 



4^ii8i9. 



Ques. 8. Required the horse power of an engine that 
would supply the city of Brooklyn with water, working 12 
hours a day, the water to be raised to a height of 50 feet ; 
the number of inhabitants = 130,000, and each person to 
use 5 gallons of water a day. 

The standard gallon of the United States weighs 8^ lbs., 
nearly. 

130000 X 5 X 8i _ 8125 X 25 lbs. 
12 X 60 ~ 27 

the pounds of water to be pumped in a minute. 
8125 X 25 



27 



X 50 = 3^6151' 



33000 

The horse power required ? 

Ques. 9. What is the horse power of an engine that 
pumps from three different levels, whose depths are 40, 70, 
and 90 fathoms respectively ; from the first 20 cubic feet 
of water are raised per minute, from the second, 10 cubic 
feet, and from the third, 35 feet, allowing i the w^ork of the 
engine to be destroyed by useless resistance ? 

T \ ^ 33000^ 

t of 33000' = *a*iOOO', 

the effective power of the engine. 

1st level, work = 62.5 X 20 X 240 = 300000^ 
2nd '' '' = 62.5 X 10 X 420 = i6*i500'. 
3rd " " = 62.5 X 35 X 540 = 1181^50^ 

n43150 



a^ooo' 



= ^9^^6. 



PRACTICAL MECHANICS. 13 

Ques. 10. There were 6000 cubic feet of water in a mine 
whose depth is 60 fathoms, when an engine of 50 horse 
power began to work the pump ; the engine continued to 
work 5 hours before the mine was cleared of the water ; 
required the number of cubic feet of water which had ran 
into the mine per hour, supposing ^ of the work of the en- 
gine to be lost by transmission ? 
4) 33000^ 

^>4^50' effective power of the 



engine a minute. It must not be forgotten that ^41150^ written 
in the contrary direction, signifies 24750 units of work in a minute 
or 24750 lbs., raised one foot high in a minute ; so that writing 
down the numbers in this peculiar manner saves much written ex* 
planation. 

^4l150' X 50 X 5 X 60 = 31^^50000, 

effective work of the engine in 5 hours. 

62.5 X 60 X 6 = *i^50, 

work in raising one cubic foot of water to a height of 60 fathoms 
= 360 feet. 

3T^^OOOO_ 

^^500 - ^^^^^-^^^ 

cubic feet of water pumped in 5 hours. 

16544.44 

6000. 

5) 10544.44 Water run in during 5 hours. 

2108.88 



Cubic feet run in in one hour. 

Ques. 11. A forge hammer weigns 300 lbs., makes 100 
lifts a minute, the perpendicular height of each lift = 2 
feet ; what is the horse power of the engine that gives pow- 
er to 20 such hammers ? 

Work of each Uft = 300 X 2 X 20 = l^OOO. 
Work in 100 lifts, that is, in one minute, 
l^OOO X 100 == l^OOOOO 



33001>' 



36 \Mi\, 



1 ! PRACTICAL MECHANICS. 

Ques. 12. All engine* of 10 horse power (lO^), raises 
4(}00 lbs. of coal from a pit 1200 feet deep in an hour, and 
also gives motion to a hammer which makes fifty lifts in a 
minute, each lift having a perpendicular height of 4 feet, 
what is the weight of the hammer ? 

Work done by the engine in one minute, 

= 33000 X 10 = 330000 . 
The units of work performed, in raising the coals, per minute = 

'^'\^ ''''- = 80000'. 
60 

Work engaged in raising the hammer per minute, 

= 330000' — 80000' = ^50000 . 

Work per minute, in raising one lb. of hammer 4 feet high, 50 
times a minute, = 

1 X 4 X 50 = *iOO^ 

— ^OO^ =1250 lbs, 

the weight of the hammer required. 



CHAPTER II. 

On the work of living agents. 



The laboring force of animals, varies very much witli the 
way in which their muscular strength is exerted ; and also, 
with the rate at wliich they labor. 

Tlie following little table shows the greatest amount of 
eflective work that a laboring man can perform under the 
diffen^nt modes in which he may exert his muscular power. 

WORK DONE BY A MAN PER MINUTE, WHEN HE 
WORKS 8 HOURS A DA.Y. 

A man in raising his own body, >4*i50 . 

A man in working at a treadmill 31M>0'. 



PRACTICAL MECHANICS. 15 

A man drawing, or pushing horizontally, Sl'lW. 

A man pushing or pulling vertically, *AS8t>'. 

A man turning a handle, *i4JOO'. 

A man working with arms and legs, as rowing, 4000^ 

UNITS OF WORK DONE BY A MAN PER MINUTE, WHEN 
HE WORKS 6 HOURS A DAY. 

A man in raising material with a pulley 1560'. 

A man in raising material with the hands, 1 4L^O', 

A man in raising material on back, returning empty,. 'I'liC. 

UNITS OF WORK DONE BY A MAN PER MINUTE, 
LABORING 10 HOURS A DAY. 

Raising materials with a wheel-barrow on ramps. . . .^^O'. 
Throwing earth to the height of 5 feet ^^O'. 

UNITS OF USEFUL WORK DONE BY A MAN IN A MINUTE, 
RAISING WATER BY DIFFERENT ENGINES, WORKING 
8 HOURS A DAY. 

With a windlass from deep w^ells, ^£560^ 

With an upright chain pump, nSO'. 

With a tread-mill, SnG'. 

With a Chinese wheel, *1161^ 

With an Archimedian screw, 1 505^ 

Raising water from a well, with rope and pail,. . . .105>4'. 



^VVORK OF BEASTS. 

The imaginary horse by which the power of steam en- 
gines are measured, is supposed to do more work than the 
general run of horses are able to perform. The work of a 
horse of average strength is about 22000 pounds raised one 
foot high in a minute, which according to my notation, is 
expressed thus : — 

A mule will perform f the work of a horse, = fiAtOftGf', 



16 PRACTICAL MECHANICS. 

4n ass will perform about } the work of a horse = 4:>400^ 

In a common pumping engine a horse of average strength 
will do only 17550 useful units of work in a minute. 

From ^iOOO' 
Take n550 

4l>450', lost in this case by friction 

and useless resistances. 

EXAMPLES. 

Question 1. How many cubic feet of clay, weighing 100 
lbs. per cubic foot, will 20 men throw 5 feet high in a day 
of 10 hours long ? 

From the foregoing table the number 4l10' is found. 
Work in a day, 

= 4:10' X 60 X 10 X 20 = 5640000'. 
Work in raising one cubic foot, 

= 100 X 5 = 500'. 

Consequently the number of cubic feet raised in one day by 20 
men, 

5640000 



500 



= 11280. 



Ques, 2. How many bricks will a man raise in a day six 
hours long to a height of 30 feet, supposing the weight of 
a cubic = 125 lbs., and 17 bricks to form a cubic foot? 

For a man working in this way, the foregoing tables 
give ll^O'. 

.-. 11^6' X 60 X 6 = 405360, 
work done by a man in 6 hours. 

125 X 30 = 3150, 
work done in raising a cubic foot 30 feet high. 

405360 ,^^^^^ 

••• 3150 ^ '"'-'''^ 
cubic feet raised in a day of 6 hours. 

Number of bricks = 108.096 X H ^ 1838. 



PRACTICAL MECHANICS. 17 

Ques. 3. How many cubic feet of water will a laborer 
working with a bucket and rope, raise from a Avell whose 
depth is 16 feet? 

In the table will be found 1054:', for a man working in 
this manner. 

t05>4' X 60 X 8 ::^ 505t)'iO, 
work done in 8 hours. 

The work done in raising a cubic foot of water 16 feet high, 

=.-. 6.25 X 16 -^ lOOO. 
Hence the cubic feet raised in eight hours by a man, 
5059'A0 



lOOO 



^ 505.92. 



Ques. 4. What weight is a man, working at a tread-mill, 
able to raise in a day of 8 hours, from a depth of 110 feet ? 

Tabular number = nSO'. 
1130' X 60 X 8 =1 830^00, 

work done in a day. 

The work required to raise one cwt. or 112 lbs. a height of 110 
feet will be, 

112 X 110 = 1^3^ ; 

the weight that may be raised by a man working at a tread-mill. 

Qices, 5. How many tons of coal would a man raise, work- 
ing with a wheel and axle, from a pit whose depth is 20 
feet, taking for granted, according to tlie tabulated state- 
ment, that a man so circumstanced, can perform 2600 units 
of work in a minute ? 

The work done in one day of 8 hours, 

^ *S600' X 60 X 8 z= l^^HOOO. 
The work to raise one ton, 

= 2240 X 20 -:zz >^4800 ; 
li>48000 ^^^^ ^ 
•*• ~^A80«- = ^^-^^^ ^^"' 
raised in a day of eight hours. 



18 PRACTICAL MECHANICS. 

Ques.{&, The ram of a pile engine weighs 7 cwt., and has 
a fall of 23 feet, how many strokes a day, will four men 
give, working at a wheel and axle ? 

Work done in one day of eight hours, 

=z ^600' X 60 X 8 X 4 = 4:»1>*£000. 

The work of one stroke, 

= 112 X t X 23 = 1803^ ; 

•• ;ftft^*> =277 strokes nearly. 

The force with which animal pull decreases with their speed ; 
the relation between the traction and speed of a horse, is express- 
ed with considerable accuracy by the following formula, 

^ =:= 250 — 41f r ; 

in which t — the traction in lbs. and r = the rate in miles per 
hour. 

Hence if the speed be 3 miles an hour, the traction will be 125 
lbs., for 

250 — 41f X 3 =r 125 lbs. 

If the speed be one mile an hour, the traction will be =: 208 J 
lbs., for 

250 — 41f X 1 = 208i lbs. 

From this formula, it is easily shown that a horse will do the 
greatest amount of work when he travels at the rate of 3 miles an 
hour, in fact 

|250 — 41f r i X r 

becomes a maximum when r = 3. 

If a body move slowly along a horizontal plane, the resistance 
to be overcome is called friction. Experiment has determined that 
this resistance on a given surface, is a fractional part of the weight 
of the- body moved, and it has also been found that any change in 
the rate of the motion of the body does not affect the resistance due 
to friction ; nor is the amount of friction altered by the extent of 
tiie rubbing surfaces. 

When a wagon or a cart is drawn along a good common road, 
the resistance of friction is about 3V of the whole load, so that a 
horse in order to draw 3120 lbs. along a road, must pull with the 
force of 104 lbs., for 

— 104 lbs. 

30 



PRACTICAL MECHANICS. 19 

A horse with this traction, according to the foregoing formula, 
moves at the rate of about 3 J miles an hour, for 

^ = 250 — 41f r, 

becomes 104 = 260 — 41f r, 

.*. r = Og. 

A carriage on a Rail road, only requires a pressure of about 3 J^ 
part of the moving weight to give it motion, or from 4 to 8 lbs. a 
ton. The fractions ^\ for common roads, and ^i^ for rail-roads are 
called "ihe coefficients of friction, as these coefficients become small- 
er, ibe rubbing surfaces become smoother. 

Let W be a weight drawn on a hor- 

izontal plane H R, by means of a \ -^v- 

weight P, attached to a cord A, go- — jj """^ — 

ing over a fixed pulley C ; then the 

weight P just necessary to draw or 

move W, along the plane, will be 

equal to the resistance of friction. In 

the case of a Rail-road, if W = 150 

tons, P will be = 900 lbs., when the "" 

coefficient of friction = j--,3__ or 6 lbs. a ton. 

It is very evident that whatever distance P descends, the weight 
W, will be drawn along the plane H R, the same distance ; hence 
the units of work done in moving W, will be the weight of P iu 
pounds multiplied by the distance in feet through which it descends, 
or the resistance of friction in pounds, multiplied by the space in feet 
over which W, is moved. 

The work of every machine is consumed by the work done, or by 
the useful work, together with the useless work, or the work des- 
troyed by the friction of the parts of the machine. I will here ex- 
plain one of the most beautiful laws of motion ; — when the work 
applied exceeds the work consumed, the redundant work goes to 
increase the speed of the parts of the machine, and at the same 
time like the fly-wheel, acts as a reservoir of work. T/iis accelera- 
tion goes on increasing until the work of the resistances + the useful 
work = the work applied ; and then the motion of the machine he- 
comes uniform. 

For example, in a Rail-road engine and train, at first the work 
of the engine exceeds the work of the resistances, and hence the 
speed of the engine goes on increasing ; but, as the speed increases, 
the work of the resistances also increases, so that ultimately the 
engine attains a nearly uniform motion, which is called the greatest 
or maximum speed, and then the work destroyed by the resistan- 
ces, will be exactly equal to the work apphed by the moving power. 



20 PRACTICAL MACHANICS. 

Ques. 7. Required the effective horse power of a loco- 
motive engine, which moves at a steady speed of 23 miles 
an hour upon a level rail, the weight of the train being 100 
tons, and the friction 5 lbs. a ton ? 

Put 2: — the reqmred horse power. 
The work of the engme per minute, 
=iz X 33000' ; 

The resistance of friction, 

= 5 X 100 = 500 lbs. 
The distance moved per minute, 

23 X 5280 _. _ . , . 

~ 60~ ' "^ ' 

Work to overcome the friction, 

= 2024 X 500 = lOl^tOOO^ 

But as the speed of the train is uniform, the work of 
the resistances will be equal to the effective work of the 
engine : 



33000 



30 KT 



Ques, 8. What is the rate in miles per hour, of a train of 
80 tons will be drawn by an engine of ^0 ^- ; the friction 
8 lbs. to the ton. 

Call X the uniform speed in miles per hour ; 
Work used in moving the train x miles, 
= 80 X 8 X 5280 x ; this 
is the work done by the engine in an hour. But the work done 
by the engine in an hour, will also be expressed by 
33000' X 70 X 60 ; 
.•. 33000' X 70 X 60 = 80 X 8 X 5280 x ; 
33000' X 70 X 60 



80 X 8 X 5280 



= 41.02 miles. 



Ques. 9. An engine of 4:8 ^-^ moves with a maximum 
speed of 33 miles an hour, on a level rail ; required the 
gross load of the train, friction 6 lbs. per ton ? 



PRACTICAL MECHANICS. 21 

Let X be the gross weight of the train in tons, then, the 
work consumed per hour in moving the train, 
= cc X 6 X 33 X 5*A80. 
Work of the engine per hour, 

^ 48 X 33000^ X GO 
When the speed is uniform or at its maximum, 
:r X 6 X 33 X 5^80 = 4^8 X 33000 X 60 ; 
48 X 33000 X 60 



6 X 30 X 528Q 



— 90 I?- tons. 



Qi^s. 10. In what time will an engine of 66 horse power 
moving a train of 200 tons, complete a journey of 100 miles, 
friction 5 lbs. per ton ; rails horizontal ? 

Work expanded in moving the train 100 miles, 
= 100 X 5280 X 200 X 5 = 5^8000000. 

Work of the engine per hour, 
^ 33000' X 66 X 60 X 430680000 
5^8000000 



130080000 



= 4.04 hours. 



Ques. 11. What work per minute will a horse perform 
when traveling at the rate of 2| miles an hour ? 

I have before shown that the traction of a horse moving at the 
rate of 2^ miles an hour, 

= 250 — 41f X 2i z= 145| lbs. 

2^ miles ^^- 13200 feet per minute, 

13200 
= -6^ = 220 feet. 

Hence the work of this horse per minute, 

= 145f X 220 =z 3'i083i^ 

Ques. 12. What load will a horse draw, traveling at 3 
miles an hour upon a plank road, whose friction is yjo of 
the whole load, the road being horizontal ? 



22 PRACTICAL MECHAl^ICS. 

The traction at this speed is 125 lbs., 

for 250 _ 41f X 3 ::= 125, 
The gross load must be 100 times this weight, 
=: 125 X 100 = 12500 lbs. 

Qms. 13. Suppose a horse to be able to perform 33000 
units of work in a minute on a horizontal road, whose fric- 
tion is sV of the whole road ; required the load and rate 
per hour ? 

Traction =(250 — 41fr) pounds. 
Space passed over in a minute, 
5280 r 



60 



r feet. 



Then ( 250 — 41fr)88r = the units of work performed by 
the horse in a minute = 3300tS^ From this quadratic equa- 
tion, the value of is found to be 3 miles ; 

.*. Traction =^ 125 lbs, 
and the load will be 

125 X 20 = 2500 lbs. 

Ques. 14. At what rate will a horse draw a ton, on a 
road whose coeflBcient of friction is -^^ ? 

2240 
Traction = =70 lbs. 

.♦. 10 = 250 — (41f) r; 

250 — 10 
... r=— — --=4.32, 

the rate is 4.32 miles an hour. 

Ques, 15. When a horse exerts a traction of 41f lbs., his 
rate of motion is 5 miles an hour, what gross load will he 
draw on a level road whose coefficient of friction is ^\, and 
what work will he perform a minute ? 

Load = 30 X 41f =r 1250 lbs. 

Distance moved in feet per minute, 

5 X 5280 ^^^. ^ 
= 440 feet. 

Work = 41| X 440 = t8333i'. 



PRACTICAL MECHANICS. 23 

Ques, 16. V>^liat must be the horse power of an engine, 
to cut 6600 square feet of planking in a day of 10 hours 
long ? 

It requires according to experiment, about 29000 units 
of Avork to saw a square foot of green oak planking, there- 
fore, 

The work in cutting 6600 square feet, 

= ^«0«0 X 6600 = l»lilOOOOO. 

... i?-«-oo=,.^„,, 

310000' „ , , 
*"•* 33000 = ^* 

Ques. 17. An engine of 24 effective horse power, cuts 
144 square feet of American live-oak in 5 minutes, how- 
many units of work are consumed in cutting a square foot ? 

The work of the engine in five minutes, 

= 33000' X 5 X 24 -^ 31>eOOOO, 

the work destroyed in cutting 144 square feet. 

Therefore the units of work expended in cutting one square foot, 

3060000 



144 



^^50o. 



CHAPTER III. 



On the moving of bodies on inclined planes^ and the raising 
of materials. 

If a surface be supposed without friction, the units of 
work performed by moving a body along it, is equal to the 
product of the weight of the body in pounds, by the verti- 
cal height in feet through which it is raised. 



24 



PRACTICAL MACHANICS. 



To move a body along the path A 
C E G I, may be imagined the same 
as to move it along and up an infinite 
number of steps, resembling- A B, 
B C, C D, D E, &c. And since fric- 
tion is neglected, there is no work 
required to move the body in a ho- 
rizontal direction, all the work is ex- 
pended in raising the body in the per- 
pendicular direction : hence the units 
of work required to raise a body from 
L to I, equals the units of work re- 
quired to move it along ths path A C E G I, the friction 
of the path being neglected. 

The principle here exDlained holds true for an inclined 
plane. 




EXAMPLES. 



Question 1. A train of 200 tons ascends an incline which 
has a rise of 5 feet in 1000, with a uniform speed of 30 
miles an hour, what is the effective power of the engine, 
the friction being 5i lbs. to the ton ? 

The pressure of a body on an inclined plane is nearly 
equal to its weight, when the inclination is small, hence 
the work due to friction, may be found bv the method ex- 
plained in the last chapter. 

Speed of train per minute =: 2640 feet. 
Weight of train in lbs. = 448000. 

5 1 . 

, rise 



1000 

of the rail in one foot. 
1 



200 



200 



X 2640 = 13.2 feet, 



the rise of the rail in 2640 feet. 

Consequently the wliole weight of the train is raised 13.2 feet 
every minute in opposition to gravity. 

Work done to gravity per minute, 

= 448000 X 13.2 11^ r>«l»«0«^ 



PRACTICAL MECHANICS. 25 

Work done to friction per minute, 

:= 200 X H X 2640 ^ *i»OiiLOOO. 

Total work of the engine per minute, 

A 

Ques, 2. A train of 330 tons, ascends an incline that has 
a rise of j in 100, what is the maximum speed with an en- 
gine of 120 horse power ; the friction of the rail 8 lbs. per 
ton? 

Let X be the speed of the train in feet per hour. 
The rise in each foot of rail, 

i^ I divided by 100 i:^ sio ; 
the rise of the rail in X feet, 

X 

"^ 500 ' 
The work due to gravity in an hour, 

= 330 X 2240 X ~r- = 14l18A x ; 

Work due to friction in an hour, 

330 X 8 X :?^ = ^64:0 x. 

It is evident that the work due to gravity in an hour, 
added to the work due to friction in an hour, must be equal 
to the work done by the engine in an hour, 

.-. 33000' X 120 X 60 := ^640 x x 14:T[8.>4 x, 
^3^600000 



>4148A 



^ 57692 feet. 



Ques. 3. A engine of 50 horse power ascends a gradient 
having a rise of f in 100 ; with a steady speed of 20 miles 
an hour, what is the w^eight of the train in tons, the fric- 
tion per ton being 8 lbs. ? 

Let X be the weight of the train in tons ; the work required to 
overcome friction. 



26 PRACTICAL ]\IECHANICS. 

= 1760 XS X x = 14l«80 x^ ; 

since a speed of 20 miles an hour =:=: 1760 feet a minute 
Rise of rail in a foot, 

= ^ divided by 100 :=:= ^^j^ ; 
Rise of the rail in 1760 feet, 

Work due to gravity, 

= 2240 XxX 13.2 =^ ^9568 x' 
Work of the engine per minute, 

= 33000' X 50 = 165000 , ^ 
Consequently, 
14.080 x^ + ^9568 x' = 1650000' 
1650000 



4l36>48 



= 37.8 tons. 



Ques. 4. A train of 100 tons descends a gradient having 
a rise of i of a foot in 100 feet, at a uniform speed of 60 
miles an hour, what is the horse power of the engine, fric- 
tion reckoned at 8 lbs. a ton ? 

60 miles an hour ^^ 5280 feet a minute ; 

Work due to friction, 

= 100 X 5280 X 8 11=: 4l^^>5lOOO^ 

Rise in one foot, 

:= i divided by 100 = 4 J©. 

Rise in 5280 feet, or in one minute, 

Work due to gravity, 

— 224000 X 13.2 — 'iUoGSOO'. 
In this case gravity acts with the engine ; 
.'. 4^*A>4«00' minus 'iOSftSOO' ::= l^GTiOO' 



•33«iW 



»8 ^.\. 



PRACTICAL MECHANICS. 27 

Ques. 5. If a horse exerts a traction of 144 lbs., what 
weight can he pull on a plank road, up a hill that has a 
rise of 3 feet in 190 feet, supposing the coefficient of fric 
tion to be 2V ? 

Work of the horse in moving over 190 feet, 
= 190 X 144=1^1360. 

The work of friction in moving x lbs. over 190 feet, 

-^ X 190 = 9.5 X, 

supposing X to be the required load ; 

The work due to gravity, when tne load is moved over 1 90 
feet = 32;; 

.•• 9 2^ + 3:^ = ^1360; 
il360 



1^.5 



2188.8 lbs. 



Ques. 6. What would be the backward pressure of a 
horse in going down hill, that has a rise of 15 feet in 369 
with a load of 2000 lbs. supposing the coefficient of friction 
to be 3V ? 

Work due to gravity in moving 2000 lbs. 369 feet, 

= 2000 X 15 = 30000 ; 
Work due to friction in moving the load 369 feet, 
2000 
= ""30^ ^ ^-^^ '"= ^4l600 ; 

From 30000 

Take ^4:600 

54lOO 

54cOO 



369 



= 14.6 lbs. 



Ques. 7. How many horses would it take to draw a load 
of 6 tons up a hill having a rise of 2^ in 100, supposino; the 
resistance of friction to be fV of the whole load, and the 
traction of eacli horse, 160 lbs. 

Let X be the number of horses. 



28 



PRACTICAL MECHANICS. 



Work due to friction in moving 6 tons over 100 feet, 
6 X 2240 



12 



X 100 = ll^OOO. 



Work of gravity in moving 6 tons over 100 feet, 

= 6 X 2240 X 21 ^ 33600. 
The work of a; horses in passing over 100 feet of this road, 
= 160 X :?: X 100 = I6OOO2; ; 
.•. 16000 X = ll^OOO + 33600, 
14^5600 



z = 



16000 



= 9.1 horses. 



11! 




— r 




c 




r ^ 


-e 






1;^ 



H 



The work in raising materials, 
having a given form, will be its 
weight multiplied by the height to 
which the centre of gravity is rais- 
ed. 

Suppose the body n 6 to be raised 
from the horizontal line H, R, . and 
the point C to be the middle or cen- 
tre of gravity of the body ; let e and 
r be points equi-distant from C. 
Now if equal weights, say for exam- 
ple 3 lbs. be placed at e and r, the 
centre of gravity of these weights 
will be at C, then 



R 



Work in raising 3 lbs. to n^ 3 X H r. 
" " '* to e = 3 X H g. 

Sum = 3 X (Hr'+Hg). 

But H r + H e :z= 2 H C ; 

.-. sum = 6 X H C. 

That is, 6 lbs. raised from H to C is the same amount of 
work as 3 lbs. raised to r, and 3 lbs. raised to e. The same 
may be proved of any two equal weights, one of them rais- 
ed as far above the centre of gravity C, as the other is be- 
low it. 



PRACTICAL MECHANICS. 29 

Ques. 8. Required the units of work in raising the mate- 
rial of a wall 22 feet long, 13 feet high, and 2i feet thick ; 
supposing the weight of a cubic foot of the material to be 
140 lbs. ? 

Contents of the wall in cubic feet, 

= 22 X 13 X 21 1=: tl5. 
Weight of the wall in pounds, 

= 115 X 140 = 10010. 
The height of the centre of gravity of the whole wall, 

= -^---^H feet ; 
Work = 10010 X 61 = 65065. 

Ques. 9. The shaft of a pit is to be sunk 120 feet deep, 
ind 6 feet in diameter ; in how many days would a man 
\Vorking with a wheel and axle, raise the material, suppos- 
..ng a cubic foot to weigh 100 lbs. ? 

Number of cubic feet in the shaft, 

= 6^ X .1854 X 120 = 3392.928 ; 

Weight of the material, 

= 3392.928 X 100 =339292.8 lbs. 

The units of work to raise this material ^f* feet, 

120 
= 339292.8 X ~- ^0351568 ; 

for it is clear that the centre of gravity of the shaft is 60 feet from 
the top. 

A man will perform ^600^ working 8 hours a day, turning a 
handle. 

Units of work in 8 hours, 

= ^600' X 60 X 8 = 1^48000. 

The number of days required, 

^0351568 



1'2A8000 



16.33. 



30 PRACTICAL MECHANICS. 

Ques. 10. Required the work in raising 3 cwt. of coals 
from a pit whose depth is 120 feet, the circumference of the 
rope being two inches, allowing the weight of one foot of 
the rope of 1 inch in circumference to be .046 lbs. ? 

Weight of one foot of rope, 

:= 2^ X .046 =^ .184 lbs. 
Weight of the whole rope, 

= .184 X 120 := 22.08 lbs. 
Weight in raising the rope, 

= 22.08 X ^1^ = 13*i>4.8. 

Work in raising the coals, . . 

= 112 X 3 X 120 := >5l03^0 ; 

Total work, 

Ques. 11. A cistern 22 feet long, 10 feet broad, and 8 
feet deep ; required the work in filling it, when the height 
of the bottom of the cistern from the water in the well is 
36 feet? 

Water in the cistern, 

= 22 X 10 X 8 X 62.5 = 110000 lbs. 

The height to which the centre of gravity of the water has to 
be raised, 



-f 36 = 40 feet. 



2 
Work = 110000 X 40 = 4400000. 

Ques, 12. The side A B of a cube of granite is 6 feet, and 
the weight of a cubic foot is 170 lbs. ; it is required to find 
the work necessary to overturn it on the edge of A ? 

The distance of the centre of gravity g, from the edge A, 

, 6' 
:=^ — -^ 4.24 feet. 

When the centre is about to fall, the centre of gravity is 
raised from r to 7i, consequently the work in overturning 



PRACTICAL MECHANICS. 



31 



the body is tlie same as raising its whole weight a perpen- 
dicular height = r n, 

.-. rn — 4.24 — 3 = 1.24. 
Work = no X 6^ X 1.24 = 4^553^.8. 

In a body of any form about to fall, c 

the centre of gravity §-, will be at n 
in the vertical line A C, the work in 
bringing the body to this position, 
is due to the vertical distance r n, 
through which the centre of gravity 
has been raised. The work necessa- 
ry to overturn any body is the true 
measure of stability. 




CHAPTER IV. 

ON THE LEVER, V/HEEL AND PULLEY, INCLINED PLANE, 
WEDGE AND SCREW. 



In this chapter I intend to treat of the transmission of 
work by simple machines, it may be observed, that the ob- 
ject of machinery, whether simple or compound, is to regu- 
late the distribution of work, or to change the direction — 
and not to increase work. 

If the parts of a machine were not subject to friction or 
any other resistances, the work given out would be exactly 
equal to the work applied. Dead matter, by its gravity, 
produces pressure, and by the intervention of mechanism 
that pressure may be increased or decreased ; but v)ork is 
peculiarly the production of active or living agents. 



32 



PRACTICAL MECHANICS. 




THE LEVER. 

Suppose two uniform bars A 
B, B C, to be suspended from 
their centres w and m, by means 
of chords attached to the points 
4 and 3 of the lever D E turn- 
ing on the fulcrum F, which 
must evidently be in the middle 
of D E = A C, and over the 
middle of A C, in order to se- 
cure equilibrium ; that is, in 
order that the two parts may 
rest horizontally as if they were 
in one piece, and suspended by the middle point S. It is 
also evident that the equilibrium will nut be destroyed if 
the bars be hung by their ends at the points 4 and 3. 

Let the weight of the bar xi. B = 6 lbs., and the weight 
of B C = 8 lbs., then A B will contain 6 units of length, 
and B C 8 units of length. It is evident from the figure, 
that F q, the distance at which A B acts from the fulcrum 
contains 4 units, and F p, the distance at which B C acts 
from the fulcrum will contain 3 units ; then since it appears 
that a weight of 6 lbs. suspended at §', balances a weight 
of 8 lbs., suspended at j9 ; therefore the following relation 
exists when equilibrium takes place ; 

6 lbs. X 4 = 8 lbs. x 3, generally 

1^ X ^F = W X F^?. 

Levers are divided into three kinds : — In the first kind 
of lever, the power and weight are on opposite sides of the 
fulcrum. 

First order of Lever. 

A B is a lever of the first or- 
der, F the fulcrum or prop, P 
the power acting at A C, and 
W the weight attached to the 
point D. If C F be twice D F, 
5 lbs. at C will balance 10 lbs. 

at D ; generally as many times as.C F is longer than F D, 

so many times will W be greater than P. 




PRACTICAL MECHANICS. 



33 



sc 



Lever of the second kind. — 
The weight is between the ful- 
crum and the power. W is the 
weight, E the fulcrum, P the 
power. 

When the lever is supposed 
to be without weight, then if 
the length of A F = a, and B 
F = _p, the power P balances 
the weight W when, 

W X i? = P X ^. 

Third kind or order of Lev- 
er. — In this case the power is 
between the fulcrum and the 
weight. P, represents the pow- S 
er, F the fulcrum, and W the 
weight. Generally if B F = 
p, and P F = g', the power P 
balances the weight W, when 



Second order of Lever. 



m B 




T 



w 



Third order of Lever. 



nB 




^ 



EXAMPLES. 



Questimi 1. In a lever of the first kind, let W = 30 lbs. 



The arm C F = 11 feet, the arm F D 

Put 2; r:=: P ia Ibs. 

Then by the equahty of moments, 
11 X:2^ — 4 X 30, 
120 



4 ; required P ? 



11 



= ioi; lbs. 



Ques. 2. A man exerts a pressure of 50 lbs. on a crowbar 
at a distance of 4 feet from the fulcrum, what weight will 
he balance at the distance of 3 inches from the fulcrum ? 

W X 3 = 50 X 48, 
.•. W-^800 lbs. 



34 PRACTlt)AL MECHANICS. 

Ques. 3. In a lever of the second kind, W = 11 lbs., 
B F = 16 inches, and A F = 100 inches : reauired P ? 

P X 100 = 11 X 16, 
.-. P== 1.76 lbs. 

Ques. 4. In a lever of the third kind, W = 40 ; B F = 
60 inches, and P F = 8 ; required P ? 

P X 8 = 40 X 60, 

.•. P z=z 300 lbs. 

Ques. 5. In a lever of the first kind, 5 and 8 lbs. are 
placed on one side of the fulcrum, at a distance of 6 and 4 
inches respectively from the fulcrum ; required the power 
P, acting at the distance of 9 inches from the fulcrum, to 
maintain equilibrium ? 

PX9=-5X6 + 8X4, 

••• ^=^='>- 

T Q Ques. 6. In a combina- 

tion of three levers of the 
first order, represented on 

^ 9 -^ — 7 3- — 8 — yt tl^^ accompanying diagram, 

A the long arms are 9, 7, 8, 
inches respectively, the short arms 2, 3. 1 inches ; If a pres- 
sure of 10 lbs. be applied at P, what is the pressure at Q 
to balance it ? 

9 X 10 = 2 X the pressure at A, 
The pressure at A =::: 45 lbs., 

7 X 45 = 3 X the pressure at B, 
Pressure at B = 105 lbs. 

105 X 8 — 1 X the pressure at Q. 
Pressure at Q = 840 lbs. 



PRACTICAL MECHANICS. 



35 




Ques. 7. Let P F and F W 

be the arms of a false balance, 
a certain weight Q weighs 16 
lbs., when put in the scale at- 
tached to the long arm, and 
only 9 lbs. when weighed in 
the opposite scale ; what is 
the true weight of Q ? 



By the equality of moments, the two following equations are ob- 
tained. 

QXWF = 9XPF; 
Q X PF= 16 X WF. 

Multiplying these equations together, and then striking out the 
common factors, 

Q^ = 9 X 16; 

.-. Q:=:12. 



THE PRINCIPLE OF WORK APPLIED TO THE LEYER. 







on 



w 



Let B F = 10 feet, F A = 2 feet, 
and P = 3 lbs. ; required W ? 

It is evident, if P be raised 5 feet, 
W will be depressed one, because 
F B is five times the length of F A ; 
consequently the work of P = 3 X 5, 

and the work of W = W X 1 ; 

.-. W X 1 = 3 X 5 ; 

.% W— 15 lbs. 

IJence the equality of moments is readily established by 
tlie principle of work, and conversely. When the motion 
is extremely small, the principle of work is termed the prin- 
cIdIc of virtual velocities. 



36 



PRACTiCAL MECHANICS. 



OF THE LEVER WHEN ITS WEIGHT IS TAKEN INTO 
ACCOUNT. 

The tendency of a unifona 
beam or lever A B, to turn 
about the fulcrum F, is just 
the same as if the whole of its 
w^eight were collected in its 
middle point, or centre of gra- 
vity C. For if the preponderating side A F, be hung from 
a cord m r, placed so that A m = n B, then this cord will 
sustain one half the weight of the beam. Now if the whole 
weight of the beam be collected at C, the centre, it would 
produce the same strain upon the cord, hence the beam acts 
as if its whole weight were collected in its middle point. 



n 

~TpB 

A 



Ques. 8. The weidit of the lever of the first kind = 10 
lbs., the length A B"= 56 inches ; A F = 40 ; C F = 36 ; 
D F == 5 inches, and W = 150 ll3S ; required P, in order 
to maintain the lever in equilibrium. In this case the 
weight of the lever acting at its centre of gravity m, co-op- 
erates with the power. 



f?z F := A F — A m^ 

= 40 — i of 56 =: 12 
Then by the equahty of moments, 

P X 86 + 10 X 12 1=5 X 150. 
P^=: n.25 lbs. 




Ques, 9. The weight 
of the lever S R, of the 
first order = 31 lbs.; S 
R = 85 inches ; S F = 
55 ; A F = 43 ; B F = 
19 ; and W = 34 lbs. 
Required P '? 



PRACTICAL MECHANICS. 



37 



P X 43 + 31 X 3| = W X 19, 



P=:^ 



34 X 19 — 31 X 3 



43 



2 



12* lbs. 



Ques. 10. The weight of a 
lever of the second order = 
8 lbs. S R = 80 inches ; S 
F = 76 ; A F = 70 inches ; 
B F = 2 inches, and P =100 
lbs. ; Required W ? 

In this case, the weight of 
the lever acts with W ; 




76 — 



80 
2 



= 36 



W X 2 -|- 8 X 36 = 100 X 70, 
,•. W = 3356 lbs. 

Ques. 11. A beam R S, whose weight is 4 cwt., is sup- 
ported by props at A and F ; a weight W, of 20 cwt. is 
placed at B ; it is required to determine the pressures on 
the props, when R S = 50 feet ; F R = 2 feet ; B F = 12 
feet : A F = 30 feet ? 

Let C be the centre of the beam, then C R = | of 50 = 25 ; 
C F = 25 — 2 1= 23 ; suppose the beam to turn on F as a ful- 
crum, the pressure on A, 

= P X 30 = 20 X 12 + 4 X 23, 
33' _ 2 



P = 



30 



= 11 



30 



cwt. 



Because the two props support the whole weight 24 cwt., 24 — 
ll/o — 12}^ cwt. the pressure on F. 



HOW TO GRADUATE THE LEVER OF A SAFETY VALVE 
OF A STEAM ENGINE. 

A F is a graduated lever of the second kind, turning up- 
on F as a centre ; V the valve of opening or closing, as the 



38 PRACTIQAL MECHANICS. 

case may oe, tlie comuiimication of steam in the boiler with 
the atmosphere ; the lever A F rests upon the pin Q V of 
the valve V, and a sliding weight W is suspended from the 
lever, enabling the engineer to place any amount of pres- 
sure on the valve ; this pressure measures the elasticity of 
the steam when it begins to escape. When steam of very 
high temperature is employed, the admission of atmospheric 
air is dangerous, and under such circumstances this valve 
may be properly termed the unsafety valve. In order to 
graduate the lever, a weight W must be found, so that 
when it is placed at A it may balance the greatest pressure 
of the steam. 

The next thing to be found, is the position of the weight 
W, to give any proposed pressure to the valve. 




EXAMPLES. 

(czues. 12. The length A F of a lever =14 inches ; the dis- 
tance F Q ==^ 2 inches ; the weight of the valve and pin = 5 
lbs.; the weight of the lever F A = 8 lbs., and the area of 
circular value in the narrowest place = 6 square inclies ; the 
it is required to find the load W, so that when it is placed 
at the extremity of the lever, the steam may have a pres- 
sure of 30 lbs. on the square inch, or 45 lbs. including the 
pressure of the atmosphere ? 

Pressure of steam on valve, 

== 6 X 30 = 180 lbs. 
Hence the effective pressure on the lever, 
= 180 — 5 = 175 lbs., 
since the weight of the lever will act at the middle point C, 
W X 14 + 8 X t = 175 X 2. 
.-. W = 21 lbs. 



PRACTICAL MECHAN1C^ 



39 



Ques. 13. At what distance from the fulcrum must the 
load be, in the last example, so that the steam may have a 
pressure of 16 lbs. over the pressure of the atmosphere ? 
16 X 6 — 5 = 91. 
Suppose D to be the required position of the load, then 
FD X 21 + 8 X t = 91 X 12; 
.•. F D = 6 inches. 

Ques. 14. In a bent lever the perpendicular A, on the 
direction of the force P is 4 inches ; and the perpendicular 
B upon the direction of W is 7 inches ; Required P when 
W = 200 lbs. ? 




P X 4 = 200 X T, 
.-. P = 350. 

Qfms. 15. A rope A D supports a uniform pole D, rest- 
ing on the ground at 0, and supporting the weight W, sus- 
pended from D ; required the tension of the rope, when A 
D = 175 feet, A = 40 feet, D = 145 feet, W = 50 
cwt., and weight of D = 10 cwt? 

In this example D may be regarded as a lever turning 
on as a centre. Draw P perpendicular to A D, and 
C R a vertical line passing through the centre of gravity 
C of the pole, then the moment of the force stretching the 



40 



PRACTICAL MACHANICS. 




rope, will be equal to the sum of the moments of W and the 
weight of the pole, that is 

Tension of cord X P = W X N + Wt. of pole X R. 

To find the perpendiculars, 

OP, DN, CR, 

in order to do this in the most simple manner, it will be 
observed, that since the three sides of the triangle A D 
are given, the area may be found by the common rule ; the 
area =* 2100 square feet ; but the area is also expressed by 

ADXOP ^_^ 



2 -^'' 


jy^, 


.*. P = 24 feet. 




But the area is also = 




^^ J ^^2100, 

2 




.•. DN = 105 feet 




0N = V 145' — 105'=: 


: 100, 


.•. OR = 50. 




.•. Tension of cord X 24 = 50 X 


10-f 10 X 50. 


Consequently the tension of the cord 




=z 229i cwt. 





THE CENTRE OF GRAVITY. 



Every heavy body is composed of an infinite number of 
particles, each of which is acted on by the force of gravity 
in a direction perpendicular to the horizon. The sum of 



PRACTICAL MACHANICS. 41 

all the parallel forces is evidently the weight of the body. 
Now there must be a point, where a single force, equal to 
the weight of the body, being applied, will produce the same 
effect, as the force of gravity acting upon the various parti- 
cles composing the body ; — this point is called the centre 
of gravity of the body. From this deflfinition it immedi- 
ately follows ; that if the centre of gravity be supported, the 
body will stand, and vice versa. That the centre of gravity 
of all symetrical, or regular bodies is in their centre of 
magnitude. That if any body, acted on by the force of 
gravity tend to turn a lever, we may regard the weight of 
the whole body to be collected in its centre of gravity. 






Ques, 16. Let A, B and C, be three bodies in the same 
right line, it is required to determine the position of their 
common centre of gravity G with respect to any assumed 
point F, when A = 6 lbs., B = 4 lbs., C = 10 lbs. ; A F 
= 10 feet, B F = 25 feet, and C F = 42 feet ? 

Let it be supposed that an inflexible rod, without weight, 
passes through the bodies, and that the system turns upon 
F as a fulcrum ; then as the whole mass may be regarded 
as acting in the point G, by the equality of moments. 

F G (6 -f 4 -f 10) == 6 X 10 -f 4 X 25 + 10 X 42, 

.•. FG X 20:= 580, 

F G = 29 feet. 

Let A, B, &c., being any number of bodies lying in the 
same horizontal plane ; it is required to determine the po- 
sition of the centre of gravity G, referred to two axes or 
lines, X, O Y, perpendicular to each other, these lines 
X and Y are termed co-ordinate axes. 

Conceive the bodies to be connected with the axis X, 
by the perpendicular rods A d, B 5, &c., then the sum of the 
moments of the bodies, tending to turn round the axis of 



42 



PRACTIQAL MECHANICS. 



O X, will be the same as the moment of the whole mass, 
collected in the centre of gravity G. From this equality 
the distance of G- from X is obtained. In precisely the 
same way, we find the distance of G from Y ; and hence 




the point G becomes known. After the same manner the 
centre of gravity may be determined, when the bodies are 
in space, by referring them to the co-ordinate planes. 

Ques. 17. The weights of three bodies A, B, C, are 10, 
17, and 9 lbs. respectively ; and their distances from X 
are 8, 16 and 28 inches, and from Y, 32, 24 and 11 inches 
respectively ; required the position of the common centre 
of gravity ? 

Let X be the distance of the centre of gravity from the 
X ; and y the distance from Y, then 

(10 -f 17 + 9) r = 10 X 8 + 17 X 16 + 9 X 28, 

.•. X = 14. 
(10 + n + 9) 7/ ::= 10 X 38 + 17 X 24 + 9 X 11, 

23 



3/ = 24 



36 



PRACTICAL MECHANICS. 



43 




WHEEL AND AXLHi. 

This useful machine is only ano- 
ther form of the lever, where the 
power is made to act with intermis- 
sion ; in its most simple form, it 
consists of a large wheel A, and a 
cylinder, or axle, B, both of which 
turn on the same axis 0. If the 
wheel A', be turned round by a pow- 
er P, applied to it, the axle B, will 
coil up the rope by which the 
weight W hangs. The lever by 
which P acts, is evidently A 0, and 
that by which W acts is B; hence 
when these weights, or pressures, 
balance each other. 

PXAO = AVXOB. 

If the wheel be displaced by a handle, then the machine 
is called a windlass. Sometimes the handle is made to turn 
a series of wheels acting on each other by means of teetn : 
when a machine has this form, it is called a crane. 

I will now consider the equilibrium, on the principle of 
work. 

When the diameter of a circle = 1, the circumference == 
3.1416. 

Wken the wheel makes 1 revolution, the axle also makes 
one. In one turn P descends a space, 

==2 X A X 3.1416, 
and W ascends a space 

= 2 X OB X 3.1416. 
The work done by P in one revolution 

=1:2 X AO X 3.1416 X P; 
and the work of W in one revolution, 

==:2 X OB X 3.1416 X W. 

It is clear that the work done in one revolution is equal to the 
work applied, friction being neglected, 



44 PRACTICAL MECHANICS. 

.-. 2 X A X 3.1416 XP:^2xOBx 3.1416 X W. 
.-. PX0A = WX0B, 

which is the relation already established, conversely assuming the 
equation of equilibrium, the principal of work is readily established. 

Ques, 18. The handle of a windlass is 18 inches, the ra- 
dius of the axle = 3 inches, and the power applied = 60 
lbs. ; what weight could be raised were there no friction ? 

Work of P in one revolution, 

= 60 X ~^^ X 3.1416. 

Work of W in one revolution, 

= W X ^^^^ X 3.1416. 

^ V 3 ^ X IS 

••• W X ^^-- X 3.1416 = 60 X ^^ X 3.1416 

...W — 360 lbs. 

Ques. 19. Required W in the last example, when the 
thickness of the cord is 1 inch, supposing that ^ of the 
work applied to be lost in friction, and the rigidity of the 
cord ? 

The cord increases the radius of the wheel I of an inch, 
hence 

Work of W in one revolution, 
_1 
1*1 

Effective work of P in one revolution, 

= ^ X GO X 3 X 3.1416, 

W X -^ X 3.1>116 = -J- X 60 X 3 X 3.1416. 

.-. W — 2T0 'hs. 



= W X i^iT X 3.1416 ; 



PRACTICAL MECHANICS. 



45 



OF COGGED OR TOOTHED WHEELS. 

Let D be a cogged wheel turning upon the same axis as 
the wheel C ; Q another cogged wheel, acted upon by the 
former and turning upon the same axis as the axle L. From 
the wheel C is suspended the weight P, and from the axle 
L the weight W ; then while P descends, the wheel and 
the cogged wheel D, will be turned from right to left, 




but as every tooth in the cog-wheel D is being turned round, 
a corresponding tooth in the cog-wheel Q will be turned 
in the contrary direction, and thus the cord L W will be 
coiled up on the axle L, and the weight W will be raised. 



Ques. 20. Let P = 120 lbs., the diameter of the wheel 
C = 3 feet, the number of teeth in D = 11, the number in 
Q = 14, the diameter of the axle L = | a foot ; required 
W, in order that equilibrium may take place ? 



For every turn of the wheel C, 11 teeth of the wheel Q 
will be turned round, therefore as many times as the num- 
ber of teeth in D can be taken out of the number in Q, so 
many turns will the cog-wheel D make while the wheel Q 
makes one. 

Let the axle L and wheel Q make one revolution, then 
the revolu-tions made by the wheel C, 



46 



PRACTICAL MECHANICS. 



11= 14 divided by = ji 
The space moved over by W, 

= i X 3.1416; 
The space moved over by P, 
14 



11 



X 3 X 3.1416; 



Work due to W = -^ X 3.14:16 X W. 



Work due to P = 3 X 3.1416 X 



14 
11 



X l^O. 



••• Since, the work due to P and W are equal during one re- 
volution of L, neglecting friction, 

~ X 3.1416 X W =: ^ X 3 X 3.1416 X 1^6, 

.•. i W= If X 3 X 120. 
W = 916 A lbs. 



COMPOUND WHEEL AND AXLE. 

In the common wheel and axle, there is a practical limit 
to the power of the machine ; for we can only increase the 
power by increasing the size of the wheel, or by decreasing 
the radius of the axle. But in the compound wheel and 
axle, a given power may be made to raise any weight what- 
ever. 




This machine consists of two axles, A and C, cut upon 



PRACTICAL MACHANICS. 47 

the same block, round which a cord coils in opposite direc- 
tions ; this cord passes round the moveable pulley D, which 
carries tbe w^eight W. Now if the handle be turned, one 
of the cords is coiled upon the large axle A, while the other 
cord is uncoiled from the small axle C, so that the rate at 
which W ascends, depends upon the difference of the cir- 
cumferences of the two axles ; and, consequently, the pow- 
er of the machine will also depend upon this difference- 
But this may be decreased to any extent, without altering 
the length of the handle. This truly ingenious contrivance 
is due to the Chinese. 

Ques, 21. Let the diameter of the axle A, = .8 feet, the 
diameter of axle 0, = .6 feet ; the length of the handle 
H A, = 3i feet ; and W, = 6090 lbs. Required P ? 

When the handle makes one turn, the cord A w411 be 
drawn up a space equal to the circumference of the axle A, 
while the cord C, will be let down a space equal to the cir- 
cumference of the axle C ; therefore the whole cord will 
be shortened a space equal to the difference of the circum- 
ferences, and because the cord is doubled, the weight W, 
will be raised a space only equal to the half of this differ- 
ence. 

H X 3 i^ie - 6 X 3 i^ie ^ ^^^ . 

Work done upon P, in one revolution, 
= T X 3.14=16 X P ; 

... t X 3.1416 XP^^^-^ ^^^^ 7 -^^^-^^^ ^X 6090 



2 



Dividing each side by 3,1416, 



... 7 P = -^ ^ '^ X 6000, 

P = 84 lbs. 



THE PULLEY. 

A pulley is a grooved wheel, turning on an axis, and 



48 



PRACTICAL MECHANICS. 



placed in a block or case. A cord passes over the groove 
of the wheel, in order to transmit the force applied, in anj 
proposed direction. There is no advantage gained by a 
single fixed pulley ; but when there are moveable pillleys, 
the weight raised will always be greater than the power 
applied ; and then the advantage depends upon the number 
of cords by which the weight is suspended. 



Ques. 25. In the annexed system of pulleys, if W 
lbs., required P, when equilibrium takes place ? 



500 




As W is suspended by two cords, c and J,' each cord will 
support 250 lbs. ; but as the cord is supposed to have a 
free motion over the wheels, the portions a, b, c, will have 
the same stretch or tension ; hence P = 250 lbs. 

Application of the Principle of Work. 

When W ascends 1 foot, the cords c and b will each be 
shortened 1 foot, and and therefore P must descend 2 feet. 
Hence the work of P 

:^P X 2; 

and the work of W :i= W X 1 ; 

.-. Px*i = wxt. 
w 



PRACTICAL MECHANICS. 



49 



Ques. 23. Let there be two moveable pulleys, each weigh- 
ing 4 lbs., the if P = 60 lbs. it is required to determine W, 
on the principle of work ? 

If P descend 4 feet, the first moveable pulley will ascend 
2 feet, and the second 1 foot. 

Consequently, the work done in raising W and the pul- 
leys = 

Wxlx4:Xlx4LX3 = Wxl*i; 
The work of P = 60 X ^ ; 

.-. W X Vi = GO X 4t. 
W =z 228 lbs. 



OF THE INCLINED PLANE. 

To find the pressure necessary to support a body on an 
inclined plane without friction. Let the weight W, be 
drawn up the inclined plane A C by means of the weight 
P, acting by a cord parallel to the plane . then whilst W 
is moved from A to C, the counterpoise weight will have 
descended a vertical space equal to A C, according to the 
principle of work. 




Work in raising W — W X C U ; 
Work due to the descent of P 

= P X AC; 

•. PXAC:=:WXCB; 

. p - ^^ w 



50 rRACJICAL MECHANICS. 

Ques, 24. The length of an inclined plane is 30 feet, the 
perpendicular height 10 feet, and the weight of the body 
W = 20 cwt., what pressure P, will be required to sustain 
the body on the plane, friction being neglected ? 

Work in raising W in opposition to gravity. 

The work of P :=^ P X SO ; 

.-. P X 30 = ^^>^iOO 

P = U6|lbs. 

Ques, 25. The length of an inclined plane is a mile or 
5280 feet, the height 38 feet, the Aveight of the body 1760 
lbs., and the friction n\i part of the weight, Avhat pressure 
will be required to move the body up the plane ? 

In this case the inclination of the plane is small, and 
hence the pressure upon it is very nearly equal to the weight 
of the body. 

.'. Pressure to overcome friction — Ve^V lbs. 
Work due to friction, =^ 

"^?* X 5^80 ^ 35^00 ; 

Work due to gravity, = 

neO X 88 = 15>4880 ; 
Work of the pressure P applied to move the body, 
= P X 5^80 ; 
.-. P X 5^80 = 35*iOO + 154880. 
.•. p z=z af) lbs. 



OF THE WEDGE, OR MOVEABLE INCLINED PLANE. 

Let ABC be a wedge, {see the last figure,) sliding on the 
horizontal plane A B, by the action of a pressure P, appli- 
ed parallel to A B, and thercl)y elevating a weight W 
which is only free to move in a vertical direction. When 



PRACTICAL MECHANICS. . 51 

the wedge begins to act, the resistance W, rests upon the 
horizontal plane A B ; but when the wedge has moved over 
a space equal to its length, W will have been elevated a 
height equal to the thickness C B of the wedge ; then if 
A B = 1.2 feet, B C = .2 feet, and W = 60 lbs., not taking 
friction into account. 

Work applied by P — P X 1.^. 

Work of W = GO X .^ ; 

.-. P X l.'i ^ GO X .'A ; 

.-. P =: 10 lbs. 

This calculation shows that the advantage of the power 
depends upon the thinness of the back of the wedge. How- 
ever, it is necessary to observe, that the astonishing power 
of the wedge, as usually employed, is equally resolvable 
into the force of impact, which is considered hereafter. 



OF THE SCREW. 



In this simple machine the pressure applied moves in a 
circle whose radius is the length of the lever, or arm of the 
screw, whilst the direction in which the work is done, is a 
right right. 

Ques, 26. The lever of a simple screw is 6 feet, the thick- 
ness of the thread = .04 feet ; If a pressure of 200 lbs. be 
applied to the lever, what pressure Avill be produced on the 
press-board ? 

Space moved over by P, in one revolution, 

— 2 X 6 X 3.1416. 
Space moved over by W, in one revolution i:=r .04 feet. 
Work of P, in one revolution, 

= 'i X G X 3.14l1« X *100 ; 
Work of W, in one revolution, 
= W X .0>i. 

.-. w X .o>4 -= 'A X « X ;^.i A l« X *AOO. 

.-. W ::^ ]h84 96 Ibsv 



52 PRAGTICAL MECHANICS. 

The equation obtained above, shows that the efficacy of 
the screw is obtained by increasing the length of the lever, 
or by decreasing the thickness of the threads. 

Ques. 27. The lever of a screw is 4 feet, and the thickness 
of the threads i inch ; required the pressure that must be 
exerted on the lever to produce a pressure of 12 tons upon 
the press-board ? 

In one revolution of the lever, the work done, 

-^ 1^ X ^'14lO X ^ = 560 ; 
for ^ of an inch 1= o of a foot. 

In one revolution of the lever the work applied, 
= P X 8 X 3.14.16 ; 
.-. P X 8 X S.liilft -^ 56t>, 
.-. P = 22.28 lbs. 

Ques, 28. The lever of a simple screw is 2 feet, what 
must be the thickness of the threads, so that a pressure of 
5000 lbs. may be produced on the press-boards, by a pres- 
sure of 100 li3S. on the handle, or lever ? 

Let X be the thickness of the thread, then, the work applied in 
one revolution, 

== 4l X 3.14L16 X 100 = 1^56.64. ; 

Work done, 

^xX 5000 = 1^56 64l ; 
.-. X ^ .^513 feet. 



OF THE COMPOUND SCREW. 

This mechanical power consists of two screws ; one of 
which screws within tlic otlier, so that wliilst the large one 
is d*escending, the small one is relatively rising w^ithin the 
large one. in consequence of this compound motion, one 
revolution of the lever, causes the press-1)oard only to de- 
scend a dirHance equal to t!io differ(?i!ce of the thickness ef 



PRACTICAL MECHANICS. 53 

the threads of the screws. Hence the advantage depends 
upon the smallness of this difference, and not upon the ab- 
solute size of the threads ; and hence the machine becomes 
analogous to the compound wheel and axle. 

Ques,29 . In a compound screw, the length of the lever 
is 2.5 feet, the distance between the threads of the large or 
hollow screw is f of an inch, and of the small one half an 
inch ; if 60 lbs. pressure be applied to the lever, what is 
the pressure on the press-board ? 

In one revolution the large screw descends f of a inch, 
but at the same time, the small screw, by turning Avithin 
the large one, ascends | an inch ; therefore the press-board 
must descend a space = 

f — ^ = ^ of an inch = ^\ of a foot. 

Work done in one revolution, 

- W X ^ ; 
Work applied in one revolution, 
60 X *i X ^.5 X 3.1>5.16 ^ »>4*iAH ; 

.-. W X -^ = »>4tfi A8. 
W = 67858.56. 

Ques, 30. If the length of the lever = 3| feet, the thick- 
ness of the thread of the larger screw ^== J of a inch, what 
must be the thickness of the thread of the smaller screw, 
so that 20 lbs. applied to the lever, may produce a pressure 
of 20 tons = 44800 lbs ? 

I of an inch == j\ of a foot. 
Let X be the thickness of the smaller screw in feet. 
.-. 'iO X ^ X 3.'14cl« ^ 4L4800 (rV — x.) 
.\ A — xz:^ .0098214 ; 
.-. .0723214 ::== z. 



54 PRACTICAL MECHANICS. 



OF THE ENDLESS SCREW. 

In this machine, the threads of a screw cut upon a cyl- 
inder, are made to act upon the teeth of a cog-wheel having 
an axle, round which a cord coils, as in the common wind- 
lass. 

This combination gives a very slow motion to tlie weight 

Ques. 31. In an endless screw, the length of the handle 
is 3 feet, the number of teeth in the cog-wheel is 24, and 
the radius of the axle | of a foot : if 36 lbs. pressure be 
applied to the liandle, what weiglit will be raised, friction 
being neglected ? 

Since the screw is fixed upon the axis of the handle, one 
turn will cause one of the teeth of the wheel to be moved 
round ; consequently the handle must make 36 turns, whilst 
the cog-wheel and axle make one turn. 

Work done in one revolution of the axie, 
z.^ W X |- X ^ X 3.1>ilO ; 
for W will be raised 4.7124 feet. 

The worK applied in one revolution of the axle, 
= 3« X 6 X 3.1>41« X ^4 ; 
.-. w X -f X ^ X 3.1>41« = 30 X « X 3.14cl6 X ^4; 
. ^ ^ 36 X 6 X 24 

W = 3456 lbs. 



THE HYDROSTATIC PRESS. 

This powerful macnine consists of two cylinders of dif- 
ferent sizes, each of which has a piston, and pressure is 
transmitted from the small piston to the large one, by 
means of water. Tlic pressure of tlie power is applied to 
the small piston by a lever of the second kind ; and the 



PRACTICAL MECHANICS. 55 

advantage of the machine depends upon the length of tliis 
lever, and the extent of surface of the large piston, compar- 
ed with that of tlie small one. 

Ques. 32. In a hydrostatic press, the surface of the pis- 
tons are 3 and 350 square inches respectively, the lever is 
30 inches long, and the piston rod is attached 2 inches from- 
the fulcrum ; If a pressure of 100 lbs. be applied to the lev- 
er, what pressure will be produced upon the large piston ? 

Let the small piston descend i of an inch, or sV part of 
a foot, then 1 cubic inch of water will be thrown into the 
large cylinder, and its piston will be raised 3^0 part of an 
inch, or 4 2V(T of a foot ; 

Work on the small piston, 

too X 30 ± ^ . ^ 

- X 36 = 4l1 3 



Work on the large piston 
W X 



>4*aoo 

w 



>i^OO 3 

W:::= 1 75000 Ibs. 



CHAPTER y. 

WORK OF STEAM AND THE STEAM ENGINE. 

If steam in the cylinder A D exert a constant, or mean, 
pressure upon the piston A B, say of 25 lbs. per square 
inch, then if a weight of 25 lbs. be placed upon every 
square inch of surface in the piston, the steam would just 
be able to move the piston with its weights or pressures 
through the length of the stroke in opposition to the weight 
or pressure ; tliereforc tlie work performed upon 1 square 



CM 



D 



56 PRACTICAL MECHANICS. 

inch of the piston in one stroke, will be 
the pressure of the steam upon 1 incli nuil- 
tiplied by the number of feet in the stroke, 
and the work upon the whole piston will 
be tlie work upon 1 inch multiplied by fhe 
number of square inches of the whole pis- 
ton. In the high pressure engine, the pres- 
sure of the atmosphere, about 15 lbs. to 
the square inch, is opposed to the pressure 
of tlie steam. Besides this, a considerable portion of the 
pressure of the steam is required to overcome the friction 
of the parts of the engine. As a mean estimate 1 lb. to the 
square inch is allowed for the friction due to the engine 
when unloaded ; and an additional friction of | the effec- 
tive pressure, or useful load, for the resistance necessary to 
overcome the friction of the loaded engine. Thus, if the 
pressure of the steam =^ 56 lbs. on the square inch, from 
which, if 15 lbs. be taken for the pressure of the air, and 1 
lb. for the resistance of the friction of the unloaded piston, 
then the remainder 40 lbs. will be taken up by the useful 
load, and the friction arising from that load, that is 

n . load ^ „ 

load 4- — — -^ 40 lbs 

8 load 
or = 40 lbs. 

.•. load 1-= 35 lbs. 

This load is termed the effective pressure of the steam. 

In the condensing eno'ine, tlie pressure of the vapor in the 
condenser, must be taken in the place of the atmospheric 
pressure, estimated at a maximum, the pressure of the va- 
por in the condenser is about 4 lbs. per square inch ; then 

^ , , load , , r ^ ' 

load -|- *— — X -|- 4 :^ total pressure oi the steam. 

In a condensing engine, let the total pressure of the steam i::^ 61 
lbs. per square inch, 



7 

.'. load -^ 49 lbs ; 
that is, the effective prcssijfe of the steaiu, in this case — 49 lbs. 



PRACTICAL MECHANICS. 57 



EXAMPLES. 



Question 1. The area of the piston of a steam engine = 
2200 square inches, the mean effective pressure of the steam 
15 lbs. per square inch, the length of the stroke 8 feet, the 
number of strokes per minute = 20 ; what is the horse 
power of the engine ? 

Work done upon 1 square inch of the piston in one stroke, 
= 15 X 8 = l*iO. 

Work upon the whole piston in one stroke, 

Work done in one minute, or in 20 strokes, 

= ^64.000 X *iO = 5^80000'. 
But the power of a horse being, 
33000. 
5^80000^ 



33000' 



160^ 



Ques. 2. The area of the piston of a high pressure engine 
is 625 square inches, the length of the stroke 6 feet, the 
pressure of the steam 40 lbs. per square inch, the number 
of strokes per minute 20 ; it is required to find the number 
of cubic feet of water which the engine will pump from a 
mine whose depth is 396 feet, making the usual allowance 
for friction and the modulus of the pump. 

The modulus of a machine, is the fraction which expres- 
ses the relation of the work done to the work applied. For 
example, if a machine should only perform one-half the 
work that is applied to it, then the modulus in this case, 
would be \ — -5. However perfectly a machine may be 
constructed, there must always be a certain amount of work 
destroyed by friction. The work that is thus destroyed 
depends upon the extent and nature of the rubbing surfaces. 
In many machines also, the power of the laboring agent 
may be exerted in a more or less efficient manner. Hence 
it is, that the modulus of one machine sometimes differs 
very much from that of another. The following table of 



58 PRACTICAL MECHANICS. 

machines for the ^raising of water is taken from Mori/is 
Mechanique Pratique. 

Modulus. 

Incline Chain pump 38 

Upright Chain pump 53 

Bucket Wheel 60 

Chinese Wheel 58 

Archimedian Screw 70 

Pumps for draining mine.- 66 

Load + -i- load + 1 + 15 = 40 ; 

8 load 
~~~ y4 • 

load= 21 lbs. 

Useful work of the engine per minute, 

= ^1 X «i5 X 6 X ^O X .66 = 1039500^ 

Work to pump one cubic feet of water a height of 396 feet, 

^ 6^ 5 X 396 = 1a>4^500, 

Consequently the number of cubic feet that will be pumped per 
minute, 

1639500' ^^ .. _ 

^^ - i^ ^ - ^ ^ — -— 42 cubic feet. 

Ques, 3. The area of the piston of a high pressure engine 
= 660 square inches, the length of stroke = 6 feet, the 
number of strokes per minute = 20 ; what must be the 
mean effective pressure of the steam, so that the engine 
may do the work of 25 horses ? 

Let X be the number of lbs. pressure on each square inch of tlie 
piston, then, the work per minute, 

-^xX 660 X 6 X *iO ; 

.-. X X 660 X 6 X taO = *i5 x 33000 ; 



PRACTICAL MECHANICS. 59 

7) 10.4166 

1.4880 

11.9046 

15.0000 

1.0000 



27.9046 lbs. total pressure of the 



Ques, 4. The area of a piston of a high pressure engine 
= 3000 square inches, the length of stroke = 11 feet, the 
number of strokes per minute = 16 ; required the mean 
pressure of the steam, so that the engine may perform the 
work of 224 horses, making the usual allowance for friction. 

Let X lbs. be the efifective pressure of the steam. 

Work per minute with z lbs. effective pressure, 

:^xX 3000 X 11 X 16 ^ 5*18000 X x ; 

The effective work per minute will also be expressed by 

3SOOO X ^^4 == T39iOOO . 

.-. 5^8000 X ^ ^ 139^000. 

.•. .r = 14 lbs, effective pressure. 

l-|-15«|_14^-»ofl4^==32 lbs. mean pressure of steam. 

Ques. 5. The length of the stroke of a high pressure en- 
gine = 8 feet, the area of the piston 1000 inches, and the 
number of strokes per minute = 20 ; what must be the 
pressure of the steam so that the engine may pump 66 cubic 
feet of water per minute from a mine whose depth is 896 
feet, making the usual allowances for friction and modulus 
of the pump ? 

The work done in one minute, 

= 6G X 6^.5 X 8»6 X 360«0«0' ; 

The modulus of the pump ^=: .66 ; 
Hence the w^ork of the engine per minute 

X .66 1= 36t>000', therefore the work per minute, 

.66 



60 



PRACTICAL MECHANICS. 



The useful work doue one ioch of the piston in one stroke, 
5600000' 



CTseful load 



^80 



8 



'i80. 



35 lbs. 



consequently the pressure of the steam 

zzz 35 _j_ -^1- ^ 15 4_ 1 zzz: 56 lbs. 

In the steam engine, the true source of work is the evap- 
orating power of the boiler. 

The magnitude of the work not only depends upon the 
quantity of water evaporated in a given time, but also up- 
on the temperature, and consequently the pressure at which 
the steam is formed. Experimental tables have been fram- 
ed, giving the relation of the volume and pressure of steam 
from a cubic foot of water ; from these tables may be found 
the volume of steam when its pressure and volume of water 
are given, and vice vei^sa. The following vrill serve as a 
specimen of such tables. 

Volume of a cubic foot of water in the form of steam at 
the corresponding 'pressures, and Temperatures, 



Total pressure 
in pounds, per 
square inch. 


Corresponding 
temperature, by 
FahrenheiVs ther- 
mometer. 


Volume of the 
Steam compared 
with the volume 
of the icater that 
has produced it. 


Total pressure 
in pounds per sq. 
inch. 


Corresponding 
temperature, by 
FahrenheiVs ther- 
mometer. 


Volume of 
Steam ; Volume 
of water = 1. 


1 


102.9 


20954 


11 


197.0 


»2222 


2 


126.1 


10907 


12 


201.3 


2050 


3 


141.0 


7455 


13 


205.3 


1903 


4 


152.3 


5695 


14 


209.0 


1777 


5 


161.4 


4624 


15 


213.0 


1669 


6 


169.2 


3901 


16 


216.4 


1572 


7 


176.0 


3380 


17 


219.6 


1487 


8 


182.0 


2985 


18 


222.6 


1410 


9 


187.4 


2676 


19 


225.6 


1342 


10 


192.4 


2427 


20 


228.3 


1280 



PRACTICAL MECHANICS. 



61 



Total pressure 
in pounds, per 
square inch. 


Corresfonding 
temperature, by 
Fahrenheit's ther- 
mometer. 


Volume of the 
Steam compared 
with the volume 
of the water that 
has produced it. 


Total pressure 
in pounds per sq. 
inch. 


Corresponding 
temperature, by 
Fahrenheit's ther- 
mometer. 


Volume of 
Steam ; Volume 
of water ^'l. 


21 


231.0 


1224 


61 


294.9 


460 


22 


233.6 


1172 


62 


295.9 


453 


23 


236.1 


1125 


63 


297.0 


447 


24 


238.4 


1082 


64 


298.1 


440 


25 


240-7 


1042 


65 


299.1 


434 


26 


243.0 


1005 


66 


300.1 


428 


27 


245.1 


971 


67 


301.2 


422 


28 


247.2 


939 


68 


302.2 


417 


29 


249.2 


909 


69 


303.2 


411 


30 


251.2 


882 


70 


304 . 2 


406 


31 


253.1 


855 


71 


305.1 


401 


32 


255.0 


831 


72 


306.1 


396 


33 


256.8 


808 


73 


307.1 


391 


34 


258.6 


786 


74 


308.0 


386 


35 


260.3 


765 


75 


308.9 


381 


36 


262.0 


746 


76 


309.9 


377 


37 


263.7 


727 


77 


310.8 


372 


38 


265.3 


710 


78 


311-7 


368 


39 


266.9 


693 


79 


312.6 


364 


40 


268.4 


677 


80 


313-5 


359 


41 


269.9 


662 


81 


314-3 


355 


42 


271.4 


647 


82 


315-2 


351 


43 


272.9 


634 


83 


316-1 


348 


44 


274.3 


620 


84 


316-9 


344 


45 


275.7 


608 


85 


317-8 


340 


46 


277.1 


596 


86 


318-6 


337 


47 


278.4 


584 


87 


319. 4 


333 


48 


279.7 


573 


88 


320.3 


330 


49 


281.0 


562 


89 


321.1 


326 


60 


282.3 


552 


90 


321.9 


323 


51 


283.6 


542 


91 


322-7 


320 


52 


284.8 


532 


92 


323-5 


317 


53 


286.0 


523 


93 


324-3 


313 


54 


287.2 


514 


94 


325-0 


310 


55 


288.4 


506 


95 


325 - 8 


307 


56 


289.6 


498 


96 


326-6 


305 


57 


290.7 


490 


97 


327-3 


302 


58 


291.9 


482 


98 


328 - 1 


299 


59 


293.0 


474 


99 


3-28 . 8 


1 296 


60 


294 . 1 


467 


100 


329.(5 


293 

1 



62 PRACTICAL MECHANICS. 

Ques. 6. In a high pressure engine, the area of the piston 
= 120 inches, the length of the stroke == 2.4 feet, the effec- 
tive evaporation of the boiler = .5 cubic feet per minute, 
the pressure of the steam in the cylinder 64 lbs. per square 
inch ; making the usual allowance for the loss due to fric- 
tion, required the useful load per square inch of piston, and 
the useful horse power ? 

f useful load z:z: 64 — 15 — 1 =^ 48. 
Useful load — 42 lbs. 

From the table it will be found that a cubic foot of water 
raised to steam of 64 lbs. pressure, has a volume of 440 
cubic feet. 

Yolume of steam evaporated per minute 

— 440 X .5 = 220 cubic feet ; 
Volume discharged at each stroke 

120 X 2.4 o K- f f 

=- — = 2 cubjc feet ; 

144 ' 

The number of strokes each minute 

The work in one stroke 

== 4c^ X I'iO X 'i.4 = 1^096. 
Work in one minute 

=3 4^0»6 X llO = 1330560 . 

1330560 ^,^, ^^ 

Horse power := *^*^4>C^i\^ '-=^ ^%3^- S^. 

Ques. 7. The area of the piston of a high pressure engine 
is 264 inches, the length of stroke 5| feet, the pressure of 
the steam in the cylinder = 72 lbs. per square inch, the 
number of strokes per minute = 36 ; required the useful 
load, the water evaporated per hour, and the useful horse 
power of the engine ? 

Lot X be the useful load, then 

:r+ y -f 15 + 1 = 72 lbs. 

1x-{-x = 392. 
2: = 49 lbs. 



PRACTICAL MECHANICS. 63 

The volume of steam discharged per raiimte, in cubic feet 

*^64 X 5 5 

---' X 36 — 363 cubic feet. 



144 

The cubic feet of steam discharged in an hour 

= 363 X 60 = 21780. 

From the table it will be found that one cubic foot of 
water yields 396 cubic feet of steam at 72 lbs. pressure. 

Consequently the cubic feet of water evaporated oer hour 

21780 

— 00. 



396 
The useful horse power of the engine 

__ ^64 X 40 X 5.5 X S« 



= IT^ 616. 



It has been found by experiment, that whatever may be 
the pressure at which steam is formed, the quantity of fuel 
necessary to evaporate a given volume of water, is nearly 
the same. Hence it follows, that it is most advantageous 
to employ steam of a high pressure. 

Ques. 8. Required the duty of the engine, in the last ex- 
ample, allowing a bushel of coals 94 lbs., to evaporate 11.5 
cubic feet of water ? 

The useful work per hour 

= ^64i X ^» X 5i X 36 X 60 == 1536^9680 ; 

which is the work of 55 cubic feet of water, or of 1 bushel of coal, 

= 1536-i9680 X ^}i^^ = 3il330*iA, 

which is termed the duty of the engine. 

Ques, 9. A train of 200 tons, moves along at a uniform 
speed of 30 miles an hour upon a level rail, the resistance 
of friction upon the rail is 5| lbs. per ton, the resistance of 
the atmosphere 33 lbs. upon the whole train when the speed 
is 10 miles an liour, the diameter of the driving wheel 7 
feet, the area of the piston 120 inches, the length of the 
stroke = | feet, and in addition to the resistances of fi'ic- 



64 PRACTICAL MECHANICS. 

tion, tlie resistance due to the blast pipe is 1| lbs. per sq. 
inch of the piston, when the speed of tlie train is 10 miles 
an hour. Required the pressure of the steam, the evapora- 
tion of the boiler, and the number of bushels of coal neces- 
sary for a journey of 500 miles, supposing that 1 bushel, or 
94 lbs. will evaporate 11| cubic feet of water? 

When the diameter of a circle = 7 feet, the circumfer- 
ence = 22 feet nearly, which is the space passed over in 
one revolution of the driving wheel. 

Total resistance to the motion of the carriages 
^ 200 X 5.5 -f ( Sy X ^^ = 1^^^ ^^s- 

The work in one revolution 

= 13in X ^^ = 301341. 

Work of 1 lb. of pressure per square inch of the piston in one 
revolution 

:=. 1 X l^O X I- X ^ = ^^O, 

for the engine has two cylinders, and each piston makes two strokes 
while the driving wheel turns once round. 

Consequently, the efifective pressure on one square inch of the 
piston, 

30^3ii ^^^^„ 

= ^^.j^Q = 42.78 lbs. 

It has been found by experiment, that the resistance of 
the blast pipe increases with the speed of the engine, 

••. Resistance due to the blast pipe 

Hence the total pressure of the steam on the piston 

42.78 
zz- 42.78-] \-iJ^]o-^i^ = 69.14 lbs. 

The number of revolutions of the driving wheel per minute 

_ 30 X 5280 

- 60X22 -^^^' 

therefore, the two oistons will make 120 X 4 — 480 strokes a 
minute. 



PRACTICAL MECHANICS. 65 

The cubic feet of steam discharged per minute will be 
■X — X 480 = 600. 



144 ^ 2 

From the table it will be found that a cubic foot of water 
produces 411 cubic feet of steam of 69.14 lbs. pressure. 

The Dumber of cubic feet of water evaporated per minute 

__ 600 • 

"~ 411 • 

As one bushel of coal evaporates 11.5 cubic feet of water, the 
number of bushels of coal used per minute 

- 600 ^ 

"" 411 X 11.5 ' 

A journey of 500 miles is performed in I65 hours, or in 
1000 minutes, at the rate of 30 miles an hour. 

Therefore the number of bushels of coal for 1000 minutes 

600 X 1000 



411 X 11.0 



= 126.95 bushels. 



Ques. 10. In a locomotive engine, the area of the piston 
is 90 inches, the length of the stroke 16 inches, the pressure 
of the steam 50 lbs., the effective evaporation of the boiler 
.7 cubic feet per minute, the diameter of the driving wheel 
5 feet ; what is the speed of the train per hour ? 

At 50 lbs pressure, 1 foot of water forms 554 cubic feet 
of steam. 

The volume of steam generated per minute 

= 554 X .^ = 38t.8. 
Cubic feet discharged in one revolution of the driving wheel 

-, . ,, 90X16 _., 
-^^ 1728 -^''- 

Therefore the number of revolutions of the drivkig wheel per 
minute 



6(> PRACTICAL MECHANICS. 

.'. The space moved over by the carriage per minute, in feet 

1:= 5 X 3.1416 X 116.3. 
Consequently tfie miles moved over per hour 
5 X 3.1416 X 116.3 X 60 



5280 



20.7. 



Ques. 11. The area of the piston of a locomotive engine 
= 80 square miles, the length of the stroke = 15 miles, the 
pressure of the steam 48 11)S. per square inch, and the diam- 
eter of the driving wlieel 5 feet ; required the effective 
evaporation of the boiler, so that the train may have a 
speed of 30 inch, an hour ; required also the effective horse 
power of the engine, and the weight of the train, taking 
the resistances to the motion of the piston as in former ex- 
amples ? 

Number of revolutions of the wheel per minute 
^ 30 X 5280 ^ 

60 X 5 X 3.1416 
Number of strokes of both pistons per minute 

= 168 X 4 --=--. 672. 

-, 1 l^^ad , ,^ , ,30 ^, .^„ 

Load -f — i" ^^ + ^ + "To" X 1. 75 ::::= 48 lbs. 

load ^ i^^XA, ^ 23.4 lbs. 

O 

Tlie effective work per minute will be 
= ^S./A X 80 X ^ X GTi 1=. 15^^i^80^ 

Hence, the effective horse power 

Now the effective work has to support a speed of 30 miles 
an hour, 2640 feet per minute, in the train in opposition to 
the resistances of friction and the atmosphere. 

Work due to the resistance of the atmosphere per minute 

(ao \2 
^^ ^ X 33 X ^6^0 ^184080 



PRACTICAL MECHANICS. 67 

Therefore the work due to friction per minute 

Work of frictiou when the train weighs x tons 
= T X ^64tO X X ; 

188400' ^„_ 

••• " ^ T X 'A040 ^ '^•'^ ^'^"^- 
Cubic feet of steam discharged each stroke 

_ 90 X 1.25 ^ 
~ 144 ' 

the cubic feet discharged in a minute 
— 80X1.25x672 
~~ 144 

From the table it will be found, that a cubic foot of 
water yields 573 cubic feet of steam, therefore, the number 
of cubic feet of water evaporated per minute 

80X1.25X672 



144 X 573 



-1^ 81 cubic feet. 



Question 12. A railroad train of 60 tons, ascends an in- 
cline that has a rise of i in 100 ; required the maximum 
speed per hour, 40 being the effective horse power of the 
engine, friction, 8 lbs. per ton ? 

Let X = - the speed of the train in feet per hour. 

the rise of the rail in x feet. 
Work due to gravity 

ftp X ^*a4 X X ^.^^ 

Work of frictiou, = GO X 8 X :?: -^ ^80 x ; 

But the work due to gravity per hour, added to the work due to 
friction per hour, must be equal to the work done by the engine in 
the same time ; 

.-. 8lfir = >^4« X S3000 X 60 izir lO^OOOOO • 



68 PRACTICAL MECHANICS. 



Hl« 



970.56 ft. .-=: 18.4 miles. 



xf the engine in the last example move this train, what 
must be the effective evaporation of the boiler, and the 
duty of the engine. 

91056 
Speed per minute, = • = 1617.6 feet. 

Number of strokes of the pistons per minute, 

-_-i^'A_ X4-412. 

The effective work of the engine per minute, 

= 33000 X >i« = 13*iOOOO'. 

Suppose y = the pounds effective pressure on each square 
inch of the piston, then the work of y lbs. effective pres- 
sure per minute 

3/ X 80 X ^- X 4cl^ ^ 13*iOOOO^ 

.•. 3/ 1=^32 lbs. 
Pressure of steam 

z.:: 32 + I X 32 4- 15 4- 1 4- ^ X 1.75 = 55.7 lbs. 

One cubic foot of water in the form of steanj, at 55.7 lbs. pres- 
sure, is 504 cubic feet. Xumber of cubic feet discharged per 
minute, in the form of steam 

80 X 15 
•*• Number of cubic feet of water evaporated per minute 

Xow this water performs 13^0000', 
.*. 11.5 cubic feet of water, or 94 lbs. of coal 

the duty of the engine. 



PRACTICAL MECHANICS. 69 

OF THE WORK DEVELOPED BY THE CONDENSATION 
OF STEAM. 

When water is raised into steam at the boiling point or 
temperature, of 212*^ Fahr., its volume is increased 1711 
times, or a cubic inch of water will very nearly form a 
cubic foot of steauL Now, if steam at this temperature be 
allowed to enter the lower part of the cylinder, (see the 
last figure), then the pressure beneath the piston will just 
counterpoise the pressure of the air upon the piston, and a 
small additional force will cause the piston to rise. If, 
then, the steam be condensed by a jet of cold water, a 
vacuum will be formed, and the piston will be pressed 
downward with the whole weight of the atmosphere rest- 
ing upon the surface of the piston. But, it has been found, 
that a perfect vacuum cannot be formed in this way, be- 
cause water gives off vapor at all temperatures. Now, if 
14.7 lbs. be taken as the mean pressure of the atmosphere, 
upon one square inch of surface, then by the condensation 
of the steam, upon an average, an effective pressure of 

U.I — i zizz 10.7 lbs. 

is established upon every square inch of the piston ; takin^ 
the temperature of the condenser at 150^, at which the cor 
responding pressure of the vapor is 4 lbs. per square inch. 
However, when the temperature of the condenser is given, 
the pressure of the vapor is easily found. 

EXAMPLES. 

Ques. 13. Required the work developed, by the conden- 
sation of a cubic foot of water, supposing 4 "lbs. to be the 
elastic^* ty of the vapor after condensation ; required also, 
the duty of the atmospheric engine using the steam in this 
manner. 

Cubic feet of vacuum foruied by condensation. 

^ 1710= 1711 — 1. 
Pressure on one square inch of the piston 
= 14.7 — 4 -=:: 10e7 lbs. 



o 



70 PRACTICAL MECHANICS. 

Suppose the area of the piston to be one square foot, the 
length of the stroke will be 1710 feet. 

Cousequently, the work of one cubic foot of water 

^ 14l>4: X lO.T X niO = ^634l168. 

And the duty or work of 94 lbs. of coal = 1 bushel, =:: duty 
of 11| cubic feet of water 

■^ *4634^68 X 11.5 = 30*il>083^0. 

Ques. 14. What must be the effective evaporation of the 
boiler of an atmospheric engine, so that the horse power 
may be 30, allowing 3 lbs. for the elasticity of the vapor 
in the condenser. 

Work of the engine per minute 

30 X 33000 -^ »«0000 . 

Work of one cubic foot of water 

= (lii.T — 3) X 144 X nio = -assioos. 

Therefore, the number of cubic feet of water evaporated per 
minute 

»»oooo 

.34. 



^881008 



OF THE USE OF STEAM WHEN USED EXPANSITELY. 

When steam is used expansively, it is allowed to enter 
the cylinder for only a part of the stroke, and then, for the 
remaining portion, the pi.^ton is moved by the expansive 
force of the steam. This is the most economical way of 
employing steam power ; for all the available work is taken 
out of the elastic vapor before it is condensed. 

When the volume of steam is inci^eased, its elasticity or 
pressure is decreased in the same ratio ; that is, if its 
volume is increased three times, its pressure will be one- 
third of what it was at first, and so on. This is Boyle^s 
law, but M. Regnault has shown, that it will not hold in 
extreme cases. 



PRACTICAL MECHANICS. 



71 



Let the steam be cut off when the^ piston is at C D, and 
let the remaining part of the stroke be divided into any 
even number of parts ; then the pressure of the steam upon 
tlic piston when it arrives at the different lines, forming the 
divisions, may be ascertained bv the law just explained. 




^' 



asis 



Let de, e 0, rn, &c., be lines containing as many units as 
there are units of pressure on the piston at the correspond- 
ing points of the stroke, then the work done from A B to 
C D, will be the area of the rectangle a d, because the work 
is the product of the pressure c d, by the space d b ; the 
woi^k done from C D to F N, will be represented by the 
units in the curved space c dfk, because the work done 
from D to g is very nearly the pressure c d. multiplied by 
the space d o, and the work done from q to t, is very nearly 
the pressure o e multiplied by o r, and so on. Now the 
smaller these intervals are taken, the more nearly will the 
areas represent the work. In order to find the area of this 
space, take the following Rule : — To the sum of the extreme 
ordioates, add four times the sum of the even ordinates, 
and twice the sum of the odd ordinates ; then this sum 
multiplied by one-third the common distance between the 
ordinates will give the area of c kfd. 



EXAMPLES. 

Ques, 15. In a condensing engine the length of the stroke 
is 5 feet, the steam is cut off at 2 feet of the stroke, the 
pressure of the steam in the cylinder 48 lb?., and the elas- 
ticity of the vapor in the condenser is 1 Ibe^. : required the 



72 PRACTICAL MECHANICS. 

work performed upon one square inch of the piston in one 
stroke ? 

Let the space through which the steam acts expansively, be 
divided into six equal parts, then each interval will be | a foot, 
by Boyle's law. 



cd = iS — 


48. lbs. 


2 

t -- — 


X 48 


38.4 lbs. 


2 


X 48 
3 


82. lbs. 


2 
sm = — 


X 48 

^2 


27.4 lbs. 


v=^ 


X 48 _ 
4 


24. lbs. 


.*=^ 


X 48 
44 


21.3 lbs. 


o 

XV — — 


X 48 _ 


19.2 lbs. 



Area oi c d y x^ or work done expansively on 1 square inch 
of the piston in one stroke 






c 



3^ + ^^*; i == 8^.9 ; 



The work done before the steam is cut off, that is, from A B to 
CD, 

= 48 X ^ = »6 ; 

Work done against the piston by the vapor in the condenser 

= >4 X 5 = *iO. 

Consequently, the total work on one inch of the piston in one 
stroke 

^ 8T.«.» + «G — *i« ^ l«3.1> 



PRACTICAL MECHANICS. 73 

Ques, 16. The area of a piston of a condensing engine 
is 1440 square inches, the length of the stroke, including 
the clearance, is 5 feet, the steam is cut oS at 1 foot of the 
cylinder, the clearance = i foot, the pressure of the steam 
= 30 lbs., the elasticity of the vapor in the condenser = 
4 lbs., the effective evaporation of the boiler .2 cubic feet 
per minute, and the resistances as before described ; re- 
quired the useful load, and the useful horse power of the 
engine ? 

By dividing the space through which the steam acts expansively 
into four equal parts, and calculating the pressures by Boyle^s law, 
the work done expansively on one inch = 

Now the space through which the piston moves before the steam 
is cut off 

= 1 — i = I foot ; 

Work done before the steam is cut off 

= SO X I = ^*i.5 ; 

Total work of steam on 1 square inch in one stroke 
= 48.66 + ^^Jt.5 = 11.16. 

11.16 

Mean pressure of the steam = — —^-r — =14.9 lbs. 

But the resistances = the mean pressure of the steam, 
.•• Load + i^ + 1 + 4 = 14.9. 

.•. Load = 8.66 lbs. 

One cubic foot of water generates 833 cubic feet of steam at 
30 lbs. pressure. 

.*. The volume of steam discharged per minute 

= .2 X 883 = n6.6. 

Volume discharged each stroke 

1440 

, . , X 1 =^ 10 cubic feet. 

14:4 



74 PRACTICAL M-ECHANICS. 

.•. The number of strokes per minute 
176.6 



10 



17.66. 



The useful work per minute will be the continued product of the 
load, the area of the piston, the length of the stroke, and the 
number of strokes per minute. 

.*. The eflTective horse power 
8.66 X 14l4.6*X ^.^5 X IT. 66 



33000' 



= 31^.1. 



Ques. 17. What must be the evaporation of the boiler 
of the engine in the last example, when the steam has a 
pressure of 48 lbs., so that the effective horse power = 40 ? 

Proceeding, as in the last example, the load is found to be 
16.6 lbs. 

Therefore, the effective wDrk in one stroke 

= 16.5 X 14.4.0 X ^.15 = 11^860. 

The effective work in one minute 

-= 40 X 33000 = ISiOOOO. 

Therefore, the number of strokes per minute 



11^860 



== 11.69. 



Since 10 cubic feet of steam is discharged in one stroke, the 
volume of steam discharged per minute = 10 X 11.69 =117 
cubic feet, nearly. One cubic foot of water forms 575 cubic feet 
of steam at 48 lbs. pressure. 

117 

.*. Cubic feet of water evaporated per minute = -^— ^ . 

o7o 

Ques. 18. Required the duty of the engine in example 
16? 

Useful work per minute 

:^3l.T X 33000 

Now, this work is done by the evaporation of .2 feet of water. 



PRACTICAL MECHANICS. 75 

.*. Work of 11.5 feet of water, or 1 bushel of coal. 

=: 31.T X 33000 X —^ = 60 millions, 

units of work, the duty of the engine. 

When water, or any body falls from a given height, the 
work which it is capable of performing, will just be that 
vfhich would be done upon it in raising it to the height 
from which it has fallen. It is no matter in what way this 
work is used ; whether the water falls into the buckets of 
an overshot wheel, or delivers its work upon the paddles 
of an undershot wheel, the laboring force of the water will 
always be equal to the work due to the height of the fall. 

Ques, 19. The breadth of a stream is 4 feet, depth 3 feet, 
mean velocity of the water 15 feet per minute, and the 
height of the fall 20 feet ; required the horse power of the 
water-wheel which does ro of the work of the water, and 
also the number of bushels of corn which the wheel will 
grind in a day of 10 hours. 

Cubic feet of water going over the fall per minute, =: 4 X 3 
X 15 = 180 ; weight = 180 X 62.5 = 11250 lbs. As the 
water descends a height of 20 feet, the work which it is capable 
to perform 

=^ ^O X 11^50 = 2^5000. 

But the wheel does ^^ of this work 

= ^^5000 X ^ = 15^500' 

Horse power = ^^^^^^^ = ^^.^T ; 

But a horse is able to grind a bushel of corn in an hour, there" 
fore, the number of bushels ground in 10 hours 

= 4.77 X 10 = 4:1.1. 



76 PRACTICAL MECHANICS. 

CHAPTER VI. 

OF ACCUMULATED WORK. 

When a body moves uniformly, the space described is, 
obviously, equal to the units of time during which the body 
moves, multiplied by the space passed over in each unit of 
time ; that is to say, the space is equal to the product of 
the time by the velocity. If, therefore, a rectangle be con- 
structed, having the units in the base equal to the units in 
the velocity, and the units in the perpendicular equal to 
the units of time ; then the units of surface in the rect- 
angle, will give the units of space moved over by the 
body. In like manner, the space described by the body, 
whose motion is uniformly accelerated, may be represented 
by the area of a trapezoid, whose parallel sides contain 
respectively the units of velocity at the commencement and 
end of the motion, and perpendicular between these sides, 
the units in the time. This proposition will readily be un- 
derstood, by observing, that the mean length of the trape- 
zoid will contain the same number of units as the velocity, 
which the body has at the middle of the time ; and this 
will be true, however small the intervals of time may be 
taken. The best illustration of accelerated motion is 
afforded in the case of falling bodies. 



FORCE OF GRAVITY. 



When bodies fall freely near the surface of the earth, 
the force of the earth's attraction being constant, communi- 
cates to them equal additions of velocity in equal intervals 
of time. Thus at the end of one second, the velocity of 
the body is 32.155 feet ; at the end of two seconds, 2 times, 
32.155 feet ; at the end of 3 seconds, 3 times, 32.155 feet ; 
at the end of 4 seconds, 4 times, 32.155 feet, and so^ on ; or 
generally, the velocity acquired by a falling body, is equal 
to the product of the time of the body's fall in seconds by 
32J feet, putting 32J for 32.155. 



PRACTICAL MECHANICS. 77 

This is expressed by the equation, 

The space described by a body in one second, will be 
half of B2f feet = IGtV feet ; because, the velocity of the 
body in the middle of the time, will be the mean velocity 
with which it moves during that time. In like manner, 
the space described by the body in 4 seconds, will be 4 
times 2 x 32^ feet ; because, 4 x 32^ feet is the velocity 
at the end of the time, and, therefore, 2 X 326- will be the 
mean velocitv, or the velocity in the middle of the time. 
But 4 times "2 X 32i = 16 X 16tV = 4^ X 16rV, that 
is, the space described by a falling body in 4 seconds, is 
equal to the square of the time, multiplied by the space 
described in one second. In the jame manner, any other 
case may be established. In general, relation of space and 
time is therefore expressed by 

^ == ^^ X 16rV. 

Ques. 1. What velocity will a falling body have at the 
end of 5 seconds ? 

32^ X 5 ri:= 1601 feet. 

Ques, 2. In what time will a body acquire a velocity of 
193 feet a second. 



193 

"32r 



=: 6 seconds. 



Ques. 3. Through what space will a body fall in 4^ 
seconds ? 

(4|)^ X 16A = 325Ufeet. 

Ques. 4. A body is thrown downward with a velocity of 
10 feet per second, how far will it descend in 6 seconds ? 

It is evident that the body will retain its motion of projection, 
although acted upon by gravity. Now the space due to the pro- 
jection rr: 6 X 10 =^ 60 feet, and this added to the space due to 
gravity will give 

60 X 6' X 16i*2 = 253 feet. 



78 pracJtical mechanics. 

Ques. 5. In what time will a body fall 193 feet ? 



By the geueral equation 
193 



12. 



16tV 
,•. ^ = 3.4 seconds. 

Ques. 6. From what height must a body fall, to ac 
quire a velocity of 96^ feet ? 

First it is necessary to find the time which the body will have 
to fall, to acquire a velocity of 96^ feet. 



96i 



■^^ 3 seconds. 



Height = 3' X 16,-^2 = 144| feet. 

Ques. 7. The velocity of a body is 84 feet. From what 
height would it have to fall, in order to attain this motion ? 



Time 


84 


; distance or space 


K- 


32^) ^ 


16, "5 


84' 


2 X 32i ■ 


g = 


32i and 


a := 


84, then 




a' 


g . 

2 


a' 



Put^ 



.•. The space fallen tlirough to acquh'e the velocity a, is equal 
to the square of (a,) divided by (2 g). In most modern works, g 
is put for the number 32^, and n is put for the circumference of a 
circle when the diameter =:^ 1. 



OF THE WORK ACCUMULATED BODY. 

When a body is in a state of motion, it will continue in 
that state, unless acted upon by some external force. But, 
in order to give this motion to the body, there must be 



PRACTICAL MECHANICS. 79 

work done upon it. Thus, we may give the velocity of 
32i feet to one pound, by raising it IGyV fe^t, and then al- 
lowing it to fall by the force of gravity. 

Iq this case, the units of work accumulated in the body will be 



Again, when a heavy fly-wheel is in rapid motion, a con- 
siderable portion of the work of the engine must have 
gone to produce this motion ; and before the engine can 
come to a state of rest, all the work accumulated in the fly, 
as well as in the other parts of the machine, must be de- 
stroyed. In this way a fly-wheel acts as a reservoir of 
work. 

In order to estimate the work in a moving body, it is 
simply necessary to consider the height from which 
it must fall to acquire the given velocity, and then 
the work will be found, by multiplying that height 
in feet by the weight op the body in lbs. ; because, 
the work expended in raising the body, to the neces- 
sary height to communicate the given velocity, must 
be the same as the work, which gravity will per- 
form upon the body in its descent 

Ques. 8. The weight of a ram is 600 lbs., and at the end 
of the blow has a velocity of 32^^ feet, what work has been 
done in raising it ? 

The ram must have fallen 16iV feet, the work done upon it 
therefore 

= 600 X 16^ = 9650. 

Ques. 9. How many units of work are accumulated in a 
body whose weight is 144 lbs., and velocity 200 feet ? 

The height from which the body must fall, to acquire the given 
gravity 

900 ' 
— ^.^"V ^. ==621.76 feet. 
2 X 32i 



80 PRACTICAL MECHANICS. 

Consequently, the work which must have been done upon the 
body 

Ques, 10. Required the work accumulated in a cannon 
ball, whose weight is 32^ lbs., and velocity of 1500 feet ? 

The height from which the ball must fall, to acquire a velocity 
of 1500 

_ 1500^ 

~2 X 32i ' 
The work accumulated in the body 

= 1500^X3^1 _ 11^5000. 

■a X 3*Ai 

This example clearly shows, that the work accumulated in a 
moving body is equal to the square of the velocity in feet per second, 
multiplied by the weight of the body in lbs., and divided by 2 X 32^. 

This very important proposition is expressed by the equation 

F»2 X W^ 



U^ 



^S 



Ques. 11. A ball weighing 20 lbs., is projected with a 
velocity of 60 feet per second, on a bowling green. What 
space will the ball move over loefore it comes to rest, allow- 
ing the friction to be yV the weight of the ball ? 

The units of work, U^ in the ball 

= 17 =^^1A^ = 1119.11. 

It is evident, that the ball will not stop until all this 
work is destroyed, that is, until the work destroyed by 
friction is equal to the accumulated work. 

Let X be the number of feet over which the ball moves before 
it comes to a state of rest, then the work destroyed by friction in 
moving the ball over x feet 

= -^ X :r = 1119.n; 



PRACTICAL MECHANICS. 1 81 



^O 



503.6 feet. 



Ques, 12. A train weighs 193 tons, and has a velocity of 
30 miles an hour when the steam is turned off ; how far 
will the train move upon a level rail, whose friction is 5J 
lbs. per ton 1 

80 miles an hour = 44 feet per second, for, 

30 X 5280 
60 X 60 ~" • 

193 tons ~ 432320 lbs, 
.•. U — ^ X 3^- "^ 130099^0. 

When the train stops, the work of friction will be equal 
to the work accumulated in the train, hence, if x be the 
number of feet the train will move after the steam is cut 
off, the work in moving the train over x feet 

= 193 X ^i XX =- U. 

.•. 193 X 5i X X = 130099^0. 

130099^0 ,,,,,, 
••• ^- 193 X5i -^^^^^fe^^> 

Ques. 13. A train weighing 60 tons, has a velocity of 40 
miles per hour when the steam is turned off, how far will 
it ascend an incline of 1 in 100, taking friction at 8 lbs. 
per ton ? 

40 miles an hour = 58 f feet a second. 

60 tons = 134400 lbs. 
Work in the train when the steam is cut off 
2\2 



(58 If X 134l400\ 



*A X 3^i 



82 PRACTICAL MECHANICS. 

put X = the number of feet the train will move after the steam is 
turned off ; then the work due to the friction 

= GO X » X a: = U^Q x , 

100:1::. :3^ = 

the rise of the rail in z feet ; 

.•• The work due to gravity 

X 



X 13^4^00 = 13>44l x; 

(58|)2 X 134L400 



100 



.% il80 X + 134c4 x = 

*X3ii 

(58 1) ^ X 134^00 
.-. X = ^— 2^ . — = 3942i feet. 

Ques. 14. Two balls each weighing 64^ lbs., are placed 
at the extremities of a horizontal arm, which gives motion 
to a screw driving a punch, as in the common stamping 
machine. The velocity given to the balls is 8 feet per 
second. What is the mean resistance opposed to the punch, 
when it is just driven through an iron plate | of an inch 
thick ? 

The weight of the two balls = 128f lbs. 

If the velocity of the balls be 8 feet, just as the punch begins 
to cut, then the units of work accumulated 

Now, if the thickness of the plate be 1 foot, the uniform resist- 
ance would obviously be 128 lbs. ; but the thickness is ^\ of a 
foot. 

1^8 

•. Mean resistance, — - — = 4096 lbs. 



PRACTICAL MECHANICS. SfrJ 

Ques, 15. A ball weighs 24 lbs., and is fired from the 
mouth of a cannon 12 feet long, with the velocity of 1000 
feet per second ; it is required to find the mean pressure 
)f the elastic vapor upon the ball? 

Let X be the pressure upon the ball during the passage through 
the barrel, then 

,•. X ::=: 31088 Ibs. mean pressure. 

Ques. 16. A carriage of 1 ton moves on a level rail with 
the speed of 8 feet per second, through what space must 
the carriage move, to have a velocity of 2 feet, supposing 
friction to be 6 lbs. per ton ? 

Work lost = — 64^ 64^—"^ *®^®- 

But this work has been taken up by friction passing over a 
^pace which put = x feet then, 

X X &^ ^089 ; 

.-. X — 348 feet. 

Ques. 17. If the carriage in the last example move over 
400 feet before it comes to a state of rest, what is the re- 
sistance of friction per ton ? 

Work accumulated in the carriage 

Work of friction ::= friction X 'iOO ; 

.'. Friction X 400 — ^^^8 ; 
Friction =:z 5.51 lbs. 



84 PRACTICAL MECHANICS. 

Ques, 18. Two weights, W and w, weigliing 5 and 3 lbs. 
respectively, are connected by a cord that goes over a fixed 
pulley, as in Atwood's machine ; through what space must 
W descend to acquire a velocity of 10 feet ? 

Since the velocity with which w ascends, is the same as the 
velocity with which W descends, then 

Work in w -- ^^ , = ^.^^a ; 

Work in *(? = — ^- ^ := 4:. 611 ; 



Total accumulated work = 1^.383. 



Let X be the space passed over by each of the weights, then 
Work of gravity on W = 5 X ^ 
Work of gravity on i^ = 3 X :t' 

Vk X X difference. 



But the work performed upon w has been yielded by the work 
of W ; therefore, the work remaining in the bodies 

= ax^=1^.383. 

.•. x=z 6.19 feet. 



OF THE CENTRE OF GYRATION, AND THE WORK IN A 
ROTATING BODY. 

Ques, 19. Two balls, A and B, are connected by a rod, 
which is made to revolve upon a centre C ; the weight of 




A is 3 lbs., and that of B is 4 lbs., the distance of A from 
the axis is 8 feet, and that of B is 5 feet ; if a point in the 



PRACTICAL MECHANICS. 85 

rod, at 1 foot from the axis C, has a velocity of 10 feet per 
second ; it is required to determine the work in the balls, 
and the point in the rod where we may suppose the weight 
of the two balls collected, so that the work may not be 
altered ? 

Yelocity A = 8 X 10 ; Velocity of B = 5 X 10 ; 
Work in A r:. ^^ ^*f f ^ ^ = ^»8.>4>4 

Work in B =^ rr^— = 155.^4: 



Total work = >5l53.88 



Let X be the distance from the axis i= C ti, then the work of 
the two balls collected at n 





>453.88 


803 X 


3 


X 


502x^ 


• 
• • 


700 a;' = 
.•. X = 


29200 
41.11 
6.46. 









C % = 6.46, the point n is called the centre, of gyration. When 
the centre of gyration in a rotating body is known, the accumu- 
lated work is readily found by the methods already given. But to 
find this centre generally, requires the aid of the Integral calculus, 
which would be foreign to the objects of this work to introduce. 
The distance of the centre of gyration from the axis, in a few of 
the most useful cases is as follows : — hi a circular wheel of uniform 
thickness, is equal to the radius of the wheel X v^ o ? ^'^ ^ ^'^^ 
revolving about its extremity, it is equal to the length of the rod 
fj I, and when it revolves about its centre, it is equal to the length 
X ^/ T 2 ; ^^^d in a plane ring, like the rim of a fly-wheel, it is 
equal to the square root of one half of the sum of the squares of 
the radii forming the ring. 

Que,s, 20. The weight of a fly-wheel is 10 cwt. This 
weight is collected in a point at a distance of 5 feet, from 



86 PRACTICAL MECHANICS. 

the axis. The wheel makes 20 revolntions per minute, the 
diameter of the axis is 2 inches, and the friction upon it 4 
of the whole weight. How many revolutions will the wheel 
make before it stops ? 

Velocity of 10 cwt. or 1120 lbs. per second 
10 X 3.1416 X 20 



60 
Work in the wheel 



= 10.47 feet. 



1908.43. 



^ 64cJ 

Circumference of the axis = -^2 X 3. 1416 = . 5236 feet. 
Work destroyed in x revolution 

= .5^36 Xx X Jr = 1908.43. 

.•, X = 22.78 revolutions before the wheel comes to rest. 

Ques. 21. The w^eight of a fly-wheel is 4 cwt., the dis- 
tance of the centre of gyration from the axis 4 feet, and 
the number of revolutions per minute 40 ; find the number 
of strokes that the fly-wheel will give a forge hammer weigh- 
ing 200 lbs., with a lift of 2 feet, friction neglected ? 

Velocity of the wheel per second = 16.755 feet. 
Work in the wheel 

Let the hammer make x lifts. Work of x lifts of the hammer 
= ^00 X "i X a; = 1954.9. 
1954.9 



400 



4.8. 



Ques. 22. The diameter of a grindstone is 4 feet, and its 
weight 300 lbs. The circumference is made to revolve with 



PRACTICAL MECHANICS. 87 

the velocity of 5 feet per second. The circumference of 
the axis is 6 inches, and the friction upon it \ of the weight. 
It is required to find the number of revolutions, which the 
stone will make when left to itself ? 

Centre of gyration from the axis = 2^1. 
To find the velocity of the centre of gyration, say, as 

The square of the velocity of the weight in feet per second 

_ 25 

— 2 ' 

Work in the stone = ^ ^ ft^^ ' — ^^'^ ' 

Let X be the number of revolutions, then the work destroyed in 
X revolutions 

— X — 2^ — X 2; =:^ o^ . o : 

.-. x = 3.109. 



OF VIS VIVA, OR LIVING FORCE. 

The vis viva of a moving body, is a force which is ex- 
pressed by the product of the quantity of matter by the 
square of the velocity. 

Let the weights of two bodies, A and B, be 4 and 7 lbs. 
respectivelv, and their velocities 5 and 9 feet per second ; 
required the accumulated work of tlie bodies m terms ot 
tl:/^ vis viva ? 

WorkinA= ^^^^"3^ 5 Work in B = ^T^^^^y 
. ' . Accumulated work in A and B 



<4 ^ ■ 3*i 



88 PRACTICAL MECHANICS. 

The latter part of the expression is called the vis viva of the 
bodies A and B. 

• • . Accumulated work n^^: i the vis viva. 



Ques. 23. What must be the weight of another body W, 
having the velocity of 8 feet per second, so that its accu- 
mulated work may be the same as that of the two bodies 
mentioned above ? 

In this case, the work in W 

»2 X w 



^ X 3*i- 



but this will be equal to the work in the bodies mentioned above ; 
therefore, by reduction, 

8^ X W = 5^ X 4 + 9^ X t 
.-. W = 10.4 lbs. 

In the above examples, if the bodies A, B, and W, are 
parts of the same machine, and if the velocity of A be in- 
creased, the velocity of B and W will obviously be increased 
in the same ratio ; hence, it follows, from the expression, 
that W will not be altered by any change in the velocity 
of the machine. 



OF THE MAXIMUM VELOCITY OF THE PISTON OF 
AN ENGINE. 

The maximum velocity will obviously take place when 
the pressure of the steam in the cylinder is equal to the 
total pressure of the resistances upon the piston ; for, so 
long as the pressure of the steam is greater than the resist- 
ances, the motion of the piston must be accelerated, then 
the work remaining in the piston, at this point of the 
stroke, will be equal to the work accumulated in the ma- 
chine. From this equality the velocity may be found. 



PRACTICAL MECHANICS. 89 

Ques. 24. The pressure of the steam is 40 lbs., the steam 
is cut off at 2 feet of the stroke, and the gross load upon 
the piston 16 lbs. per square inch. At what point in the 
stroke will the velocity of the piston be the greatest ? 

Let X = the length of the stroke, then according to Boyle^s law, 
the pressure at this point 

2 X 40 



but the velocity of the piston is greatest when the pressure i= 16 
lbs. per square inch, 



16 


=z 


2 


X 


40 




X 






. 


% : 


^z. 


5 feet. 



Ques. 25. The length of the stroke in a condensing en- 
gine is 10 feet, the pressure of the steam 30 lbs., and the 
steam is cut off at 2 feet of the stroke. Required the gross 
load upon each inch, and the point at which the velocity 
of the piston is greatest ? 

Dividing the space through which the steam acts expansively 
into four equal parts, and finding the pressures by Boyle's law, the 
total work of the steam upon 1 inch of the piston 



I 30 + 6 + 4. (15 + 1.5) + ^ X lO } + 30 



-^ •j;^i> + l> + >5^(15 + ^.5) + -J& XlM^+;-5il X 
^ = 15T.333. 



. • . Mean pressure of the steam, or gross load 
157.333 



10 



= 15.73 lbs. 



Since the mean pressure of the steam is equal to the total re- 
sistances which the steam has to overcome throughout the stroke, 
the maximum velocity will take place when the steam, expanding 
itself, attains a pressure equal to 15.73 lbs. ; hence, by Boyle's 
law, 

30 X 2 = 15.73 X ^, 

15.73 : 30 : : 2 : 2:: 



90 PRACTICAL MECHANICS. 

X being the point in the stroke when the velocity is the great- 
est ; 

.-. X = 3.813 feet. 

Ques. 26. If the area of the piston in the last example 
= 4000 square inches, and the weight of the mass moved, 
calculated to have the same motion as the piston, = 50000 
lbs. ; what will be the maximum velocity of the piston ? 

Having determined in example 25, the point where the maxi- 
mum velocity takes place, namely, 3.813 feet, or 4 feet nearly, 
from the end of the cylinder, the first thing to be done is to find 
tlie work of the steam up to this point. Total work upon the 
whole piston to the point where the maximum velocity takes place 



m 



30 + 15 + 4c c^^ -f n.i>4i; + ta X ^o 

+ 30 X ^"1 X 4000 = >4063^8 ; 

the steam acting expansively over 2 feet of the stroke, and divid- 
ing this space into four equal parts. Now, the excess of this work 
over the work which is expended in moving the resistances, will 
leave us the work which is accumulated in the piston. But the 
work done upon the resistances 

= 15.133 X 4l©00 = ^5n33. 

Work accumulated in the piston 

= 4c©6318 — ^5n33 == 154^6>45. 

Put V = the maximum velocity of the piston in feet per second, 
then 

^•2 X 50000 



e^i 



= 15^1645. 



14.1 feet. 



In the same manner the velocity may be found for any part of 
tlie stroke. The weight of the mass moved in any engine referred 
to the piston, may be found as in example 23. The mechanical 
principles evolved ia the last four examples are worthy of particu- 
lar attention. 



PRACTICAL MECHANICS. 91 



VELOCITY ACQUIRED BY A BODY DESCENDING AN 
INCLINED PLANE. 

When a body freely descends an inclined plane by the 
force of gravity alone, the work accumulated in the body, 
will be simply that which is due to the perpendicular ele- 
vation, without any regard to the angle or curve of the 
plane. Hence, the velocity of the body will be that which 
it would acquire by falling freely through the perpendicular 
height. 

Ques, 27. What velocity will a body acquire in descend- 
ing an inclined plane, whose perpendicular height is 579 
feet? 

t'' X 16yV = 519 

^^ = 36 

^ mzi 6. seconds. 

Consequently, the velocity acquired at the end 

= 6 X 32| = 193 feet. 

When the friction is taken into account, it will be more con- 
venient to employ the usual expression for accumulated work. 

Qiies, 28. A train of 10 tons descends an incline of 200 
feet, having a total rise of 3 feet ; what will be the veloc- 
ity acquired by the train, supposing the friction to be 8 lbs. 
per ton. 

Work in the train 

= lO X ^^>40 X 3 — lO X » X ^OO ^ 51^00. 

Let V = the velocity. 



G4, r— = siioo, 



3 

V = 12.1 feet. 



92 PRAdTTCAL MECHANICS. 



TO FIND THE FRICTION UPON AN AXIS BY EXPERIMENT. 

Ques. 29. The fly-wheel in example 20, was observed to 
make 28 revolutions before it came to a state of rest ; what 
part of the weight of the wheel is the friction upon the 
axis ? 

Let z be put for this proportional part, then the work destroyed 
by the frictioD 

-^ Z X ll^O X .5^36 X ^8 ; 

but this work must be equal to the work accumulated in the wheel, 
hence, 

Z X 164^^0. 00« = 1909.4.3. 

.•. z -= .116. 



OF THE RESISTANCE OF FLUIDS. 

The resistance of a fluid to the motion of a body is occa- 
sioned by the force necessary to displace that fluid. Now 
the fluid displaced must have the same motion given to it 
as that of the moving body ; hence the work destroyed by 
the fluid will be equal to the accumulated work in the fluid. 
Let, for example, the front of the body, presented to the 
fluid, contain 2 square feet, tlie weiglit of a cubic foot of 
the fluid 62| lbs., and the velocity of the body 9 feet per 
second, — then the weight of the fluid displaced every 
second = 9 X 2 X 62 1 lbs., but this weight has a velocity 
of 9 feet given to it ; therefore, the work expended in the 
displacement 

__ 02 X » X ^ X e^i 

Now this work has been destroyed, while the body has moved 
the space of 9 feet ; if 3/ be put for the resistance in lbs. this work 
will also be represented by 

y X » ; 



PRACTICAL MECHANICS. 93 

» X \ 



« t>2 X« X 'i X «*ii 

J' ^ ^ ^ o¥T 



y 



«3 X *a X 6^i 



tt>4i 



Hence, the resistance increases with the square of the velocity, 
as well as with the extent of surface presented to the fluid. In 
extreme velocities this law does not hold strictly true. It appears 
also from certain recent railroad experiments, that the resistance 
of the atmosphere to the motion of the train, depends chiefly upon 
the length of the train, and not so much upon the extent of the 
frontage of the carriages. 



OP THE EFFLUX OF FLUIDS. 



If an aperture be made in the side of a vessel, kept full 
of water, then the accumulated work of the fluid, as it is 
being discharged, will just be equal to the work which 
gravity would perform upon it, while descending a space 
equal to the perpendicular depth of the orifice. Hence it 
follows, that the velocity of the discharge is equal to the 
velocity which a body would acquire in falling freely 
through a space equal to the perpendicular depth of the 
orifice. 

Ques. 30. The cross sectional area of a pipe is 36 square 
inches, how many cubic feet will it discharge per minute 
from a vessel kept constantly full, the depth of the dis- 
charge pipe being 16 yV feet 1 

Let V be the velocity of the discharge per second, and ic the 
weight of a fluid particle in lbs., then, the accumulated work in 
the particle 

Work of the particle in descending IGy'g f^et 
— 16 j\ X W ; 



94 PRACTICAL MECHANICS. 



04ti 



= 1$5 yV X w ; 



!; = a/ 16 yV X 64^' :=.-. 32 1 feet, 
iubic feet 

321 X 60=::482i. 



Discharge per minute, in cubic feet 
36 



144 

Ques. 31. If the section of the cistern, in the last ex- 
ample be 8 square feet, and a pressure of 10000 lbs, be 
applied to a piston fitting the cistern and in contact with 
the water ; what will be the velocity of the discharge ? 

It is evident, that the effect of the pressure of the piston upon 
the efflux, will be equivalent to an additional column of fluid pro- 
ducing the given pressure on the piston. 

Height of additional column 

10000 ^^ ^ , 

- FxW = ^' ''''- 

Height of the total column of fluid 

^16-,V + 20 = 36J^; 

then proceeding as in the last example, the velocity r, will be 
found 

= V 64 i X 36 J^ = 48 1 nearly. 



OF THE WORK PERFORMED BY A JET OF STEAM. 

When steam issues from a nozzle, its elasticity is nearly 
the same as that of the surrounding atmosphere ; and the 
volume of steam at the nozzle will therefore be 1711 times 
that of the water from which it is raised. The motion of 
the particles of the steam at the nozzle, is due to the work 
of the expansion of the steam. 

Ques. 32. Steam is discharged from a nozzle, whose area 
is 8 square inches, with the velocity of 600 feet per second ; 



PRACTICAL MECHANICS. 95 

how many horse power would a wheel perform, that takes 
up all the work in the steam*? 

Cubic feet discharged per second 

144 

Weight of steam discharged per second 
12.5 X 62.5 



1711 

Accumulated work per second 
600' X .4566 



= .4566 lbs. 



64i 



= ^S58*A 



^58^ X 60 , . . 

Horse power = 330OO " — ^ '^' 

Ques, 33. If 9 cubic feet of water be evaporated per hour, 
when the area of the nozzle is 1 square inch ; how many 
horse power will the wheel haye ? 

Cubic feet of steam discharged peir second 

— ^^11 X 9 ^ 
"" 60 X 60 ' 

Velocity of the steam per second 

3600 ' 

Weight of the steam discharged per second 

3600 ^ 

Accumulated work per second 

^ res./. X 9;j^x :O n x o ^^.^^ ^^s; 

From this expression it appears, that the work performed by 
steam in this manner, increases with the cube of the water evapor- 
ated ; 



96 PRACTICAL MECHANICS. 

Horse power ^ ^^^^ = !>.. .Q. 

In these calculations the co-efficient of efflux has been neglected. 




OF poncelet's water wheel. 

In the common under-shot water wheel, the paddles are 
flat, whereas, in Poncelet's wheel they have a curved shape, 

A B ; so that the direction of 
3 the curve at A, where the water 

meets the paddle, is the same as 
the direction of the stream. 

By this ingenious contrivance, 
the water rolls up the curved ir?- 
cline A B, without meeting with 
any sudden obstruction calcu- 
lated to occasion a loss of work. 
The channel has a depression at the point where the water 
falls from the paddles. Let Y be the velocity of the steam, 
and V that of the wheel, then since the point A of the pad- 
dle is moving away from the stream, the water will flow 
upon the paddle with the velocity V — v, and will continue 
to rim up the curved line until it has lost its motion, it will 
then descend, acquiring in its descent the same velocity as 
that which it had in its ascent, but in a contrary direction. 
If the wheel were not to move during its return, V — v 
would be exactly the velocity of discharge from right to 
left, but the paddle is moving the water from left to right 
with the velocity v, therefore, the absolute velocity of the 
water upon leaving the paddle will be V — v — v ==Y — 

Now all the work will have been taken out of the water, 
when its motion upon leaving the paddle is nothing, that is, 
when V — 2v = o, 

or V :== 2 V. 

In this case the water having lost all its motion, will 
simply drop from the paddle, and the work done upon the 
wheel will be equal to the work accumulated in the water 



PRACTICAL MECHANICS 97 

of the stream. Moreover it appears, that the maximum 
condition is fulfilled when the velocity of the stream is 
double that of the wlieel. However, the distinguished in- 
ventor states, that, in practice, the velocity of the water, 
in order to procure it maximum effect, ought to be 2^ times 
that of the wheel, and that the modulus of the wheel is 
about y\. This wheel, other things being the same, will 
perform about twice the work of the common under-shot 
wheel. 

When there is motion in the water after leaving the 
wheel, the work accumulated in the water must be calcu- 
lated, and subtracted from the whole work originally in it, 
in order to obtain the work done upon the wheel. 






Ques. 34. Suppose 193 cubic feet of water flow upon the 
paddles of a Poncelet wheel per second, with the velocity 
of 8 feet., when the wheel moves at the rate of 3 feet per 
second ; required the horse power of the wheel ? 

This is obviously a case of maximum work. Weight of the 
water flowing per secoud 

= 193 X 62.5 = 12062.5 lbs. 

Work per second 

Horse power =^ *^*^Ot^O ~ .818. 

No account is taken of friction and other resistances. 

Ques, 35. Required the same as in the preceding ex- 
ample, when the quantity of water is 386 cubic feet per 
second, the velocity of the stream 8 feet, and of the wheel 
2 feet ? 



98 



PRACTICAL MECHANICS. 



Hiere the velocity with which the water leaves the paddles 
= 8 — 2X2 = 4 feet. 

The work done upon the wheel per second 
= 18000. 

Weight of the water = 386 X 62.5 lbs. 
•i X r386 X 6^.5J 



»-i 



X f8 — *1) X "^a = 18000'^ 



Horse power = ^.^^^^ = 3^^" "^^ 



. THE FLY-WHEEL. 

To find the position of the crank corresponding to its maxi- 
mum and minimum velocity in a single acting engine. 

Let P, and D, be the required positions of the crank, 
and let P be supposed to be the constant pressure of the 

connecting rod, acting always in 
a vertical line. 

Let Q be the constant resist- 
ance, acting at 1 foot from the 
axis of the fly-w^heel, equivalent 
to the work of the engine. The 
motion will be accelerated from 
P to D. This acceleration will 
commence when the moving pres- 
sure is equal to the resisting 
pressure, and will cease under 
the same condition. The former 
will correspond to the position 
of minimum, the latter to that of 
the maximum velocity. Hence, 
at these two points, the moment of P must be equal to the 
moment of Q, and the point D will be as much below the 
horizontal line V, as the point P is above it. 




PRACTICAL MECHANICS. 99 

.-. P X B = Q X 1 foot, -= Q X 1. (y)^ 

Again, by the equality of work, and putting r = Y ; and n 
= 3.1416, the circumference of a circle when the diameter is 
unity. 

Work of P in one revolution = *i r JP ; 

Work of Q in one revolution = *i rr ^ ; 

.-. ^ r Jf» = 'i ^ ^ ; (z). 

Dividing equation (y) by equation (z)^ 

OB 1 1 ^,^^ 

then = — = = .3183 ; 

r n 3.1416 

this is the cosine of the angle P V ; hence from the tables, 
an^leP V=n°..2t'. 



TO FIND THE DIMENSIONS OF THE FLY-WHEEL. 

Let d and p be the maximum and minimum velocities of 
the wheel at the distance of 1 foot from the axis ; w, the 
weight of the wheel, and k the distance of the centre of 
gyration from the axis. 

Work of P from P to D 

^PXP n^'LrP sin. ^1°..^T =:^^rPX .t>48 ; 

Work of Q from P to D 

-= 36«^ = ^ r JP X .3968, by sub- 
stituting the value of Q X 2 tt given in equation (z). 

Now the difference of these will give the work that goes on in- 
creasing the speed of the wheel between the points P and D, that 
is, work going into the wheel between P and D = 

"iir P y .»>i8 — *i r P X .39«8 :::::z r P X LIO*^ ; 



100 PRAtJTICAL MECHANICS. 



Accumulated work at P = 
Accumulated work at D :=: 






^i,g 



Difference = ^^ ^ x (d^ — p^), 



work gained from P to D. 

But this must be equal to the work before found 
fc3 t^ 



*ig. 



(d^ —p^) = r P X l.lOta^ ; (1). 



Let V be the mean velocity of the wheel at 1 foot from the axis, 
and let the velocities d and p, at each of the extremes, differ from 
the mean by the Tith part ; then, 

d = V -A and p z= v — — : 

n n 

.-. d^-p^ = --; (2). 

Let 17 be the work of the engine, and JV the number of strokes 
performed per minute ; 

then V = 2 X 3.1416 X -^ = -10472 X iV^; (3), 

.. U^'SirPJr .-. rP=-^; (4). 

Substituting the values given in equations (2), (3), and (4) in 
equation (1), and reducing the given quantities, there will be ob- 
tained finally 

"^^ 1^^ X80H.^. (5). 

After the same manner may be derived an equation, ex- 
pressing the relation of the elements for the double acting 
engine. 



PRACTICAL MECHANICS. 



101 



Ques. 36. In a single acting engine of 30 horse power, 
the number of strokes performed by the piston = 20 per 
minute, the distance of the centre of gyration of the fly- 
wheel from the axis = 10 feet ; it is required to find the 
weight of the wheel, so that the velocity of each revolu- 
tion may not vary by more than ^ from the mean ? 

n = 5y 17 = SO X 33000 ; ^ = 10 ; iV^ = 20. 

5 X 990000 



W 



lO'^ X 20^ 



X 808.2 = 500014 lbs. 



CHAPTER VIL 

OF THE EQUILIBRIUM OF FORCES AND PRESSURES. 

Let the board A W R, be placed upon three balls roll- 
ing freely upon a horizontal table. Take any three points, 
B, Q, S, in the surface of the board, and let the cords AB, 




D Q, R S, be attached to the points. These cords pass 
over pulleys, and have different weights hanging at their 
extremities. Under this arrano^ement the board, acted 



102 PRACTICAL MECHANICS. 

upon by three forces, will roll upon the table until it comes 
to a position where the three forces destroy each otlier. 
When this position has been attained, the following import- 
ant relations will be observed, between the direction and 
magnitude of the forces. 

1. The directions of the forces will intersect, or meet, in 
the same point 0. 

2. From a scale of equal parts, take off r, equal to the 
units of pounds in the weight or force, drawing the cord 
Q D, and in like manner from the same scale of equal parts 
take off v, equal to the units of pounds in the weight, or 
force, drawing the cord A B, then if upon the two lines 
r, V, a parallelogram r ev he constructed, the diag- 
onal e will be the direction of the third force acting by 
the cord R S, and the units of length in the diagonal e, 
will giye the units of pounds in tlie weight, or force, acting 
by the cord R S. The forces represented by the sides r, 
and V, are called the component forces, and the force just 
equivalent to these two and represented by e, is termed 
the resultant. 

3. Instead of three forces being applied to the board, let 
there by any number. Let P be any point taken in the 
board, and from this point let fall the perpendiculars P 7i, 
P 5, P ^, &c., on the directions of the forces, or if neces- 
sary, on the directions of the forces produced ; then the 
units of length in any of these perpendiculars, multiplied 
by the units of pounds in the corresponding force, will be 
the moment of that force, tending to turn the board upon 
the point P ; and then the principle of the equality of mo- 
ments will be obtained. 

FnXOB-^FtX r-^^F s X e. 
EXAMPLES. 

Ques. 1. Let A B, be a platform with a load C upon it, 
supported by a chain AD; it is required to determine by 
construction the tension of the chain, and also the amount 
and direction of the pressure upon the liinge at B, the 
wciglit of the platform being neglected ? 



PRACTICAL MECHANICS. 



103 




CONSTRUCTION. 

Through the point C, draw the vertical line C 0, inter- 
secting the direction of the chain in the point 0, join B 
and 0^ and from a scale of equal parts set off n equal to 
the units in the weight placed at C ; draw r n parallel to 
A D, and r e parallel to C 0, then n r e will be the parallel- 
ogram of pressures. Because, is the intersection of two 
of flic forces or pressures ; therefore, the direction of the 
third force must pass through ; but this third force must 
also pass through B. Hence it follows, that B is the 
direction of the pressure upon the hinge, and in like man- 
ner e, will give the units of pressure tending to break 
the chain. 

dues. 2. A pole, D, supported by a cord A D, carries 
a weight W ; it is required to find the tension of the cord, 
and the pressure on the pole ? 




CONSTRUCTION. 

On the line D N, mark off D n equal to the units of 



104 



PRACTTOAL MECHANICS. 



weight ill W, draw 7i7n parallel to DA, intersecting D 
m m, and from m draw m e, parallel to B n ; then Bnme 
will be the parallelogram of pressures ; therefore, the units 
m De will give the tension of the cord, and the units in 
D m will give the pressure upon the pole. 

Ques. 3. A beam, F B, is supported by a cord FA; it is 
required to determine by construction, the direction and 
tension of the cord B H, so that the beam may not change 
its position ? 

COXSTRUCTION. 

Through the center of gravity, 
C, draw the vertical line C K, 
produce A F, until it intersect 
this vertical line in the point K, 
join K and B, then K B produced, 
will ^ive the direction of the cord. 
Take K L, equal to the units in 
the weight of the beam, and con- 
struct the parallelogram of pres- 
sures, K N L V, then the units in 
K V, will give the tension of the 
cord H B. 

Ques. 4. A gate, A H, is supported by a pin turning in 
a socket at 0, and prevented from falling in the direction 
A D by a hook and loop at A ; it is required to determine 
the amount and direction of the pressure upon the pin 0, 
and the force tending to draw the hook A from the wall ? ' 





CONSTRUCTION. 

Through the middle, or center 
of gravity of the gate draw the 
vertical line D C, join the points 
D and 0, then D will be the 
direction of the pressure upon 
the pin. Take D B equal to the 
units in the weight of the gate, 
and construct the parallelogram 
of pressures F D B Q, then the 



PRACTICAL MECHANICS. 



105 



units in D Q will be the pressure upon the pin, and the 
units in D F will be the force tending to draw the hook. 



TO FIND WHETHER A PILLAR WILL STAND OR FALL WHEN 
ACTED UPON BY A GIVEN PRESSURE. 

Ques, 5. Let P be the direction and amount of the pres- 
sure, tending to turn the pillar upon the edge as a cen- 
tre ; and A Y a vertical line passing through the centre 
of gravity of the pillar, intersecting the line P, produced 
in the point A ; from a scale of equal parts, take A C equal 
to the units in the pressure P, and from the same scale take 
A B, equal to the units of weight in tjie pillar ; construct 
tlie parallelogram of forces AB D C, then A D will be the 
amount and direction of the single force tending to over- 




turn the pillar. If A D produced, intersect the base within 
the edge 0, the pillar will stand ; and on the contrary, if 
the point of intersection Q,fall without the base, the pillar 
will fall ; but if it intersect at the edge 0, then the pillar 
will just be upon the point of overturning ; the point Q is 
called the point of resistance. 

A structure will be more or less stable, according as the 
point of resistance^ Q, is more or less distant from the edge 
0. Hence, the modulus of stability, may be defined to be, 
the ratio that Q bears to S. Thus, if Q were in the 
middle between and S, the modulus would be = -| ; if Q 



106 



PRACTICAL MECHANICS. 



were at S, the modulus would be 1, or the most stable pos- 
sible, under the given conditions ; and if Q were at 0, the 
modulus would be zero, that is, the structure would be on 
the point of overturning. Marshal Vauban considered, 
that the modulus of a good structure ought to be about I. 

PRESSURE OF ROOFS. 

Ques. 6. Let C A, and Q A, be the rafters of a roof rest- 
ing upon the side walls C and Q. It is required to deter- 
mine the thrust on the points C and Q, the roof being with- 
out a tie beam ? 

CONSTRUCTION. ' 

Let it be supposed, that the weight of the roof is equally 
distributed over the surface, and that a foot of length of 
this roof acts upon a foot of length of the side walls. 




Find the weight of each side of the roof one foot long, 
and let a weight W, be suspended from A equal to half tlie 
sum of these weights, then one half of the whole weight 
upon the rafter, C A, will act perpendicularly upon the 
wall at C, the other half acting in the weiglit W ; and in 
like manner one lialf of the whole weight upon the rafter, 
Q A, will act perpendicularly upon the wall at Q, the other 
half acting in the weight W ; now take A, equal to the 
units of weight in W, and construct the parallelogram of 
pressures A e, n, then A e will give the thrust upon the 
rafter A C, and A ii the thrust upon the rafter A Q. 



VRACTICAL MECHANICS. 



107 



Ques. 7. To find the tension of the tie beam Q C, or of a 
cord connecting 'he feet of the rafters ? 

CONSTRUCTION. 

Take C a = A e, the thurst upon the rafter AC, draw 
a P perpendicular to C Q, and construct the parallelogram 
at C P then, the pressure C a is equivalent to the two pres- 
sures C t, and C P ; now the latter pressure gives the teu-_ 
sion of the cord which would be produced by the thurst of 
the rafter C A ; but the rafter A Q, will P^o^^^ce an equa 
and opposite tension, therefore, the units in 2 X C F, will 
give the units of tension of the cord C (.^. 

Ques. 8. To find the amonnt and direction of the pressure 
of the roof tending to overturn the side walls ? 

It has been shown, that besides C A, the thrust upon the 
rafter, we have a vertical pressure upon the wall, arising 
from one half the weight borne by the rafter, lake ^, 
equal to the units in this vertical pressure and construct 
the parallelogram Caqt, then the diagonal C g, will give 
the amount and direction of the pressure of the roof tend- 
ing to overturn the wall C. 

Ques. 9. Pressure of fluids on Embankments. 

Let H B, be the side of a vessel filled with water, then 
the pressure of the fluid upon the point P, which may be 




anv point in the side, will be due to the perpendicular depth 
H P. That is, in other language, the pressure is in pro- 
portion to the depth. The reason of this is, since gravity 



108 PRACTJCAL MECHANICS. 

acts upon all the particles of the fluid, each particle presses 
on that which is next below it, and tlien, from the peculiar 
property of a fluid, this pressure is transmitted equally in 
every direction. 

In the base B C produced, take B A, equal to B H, the 
perpendicular dei:)th of the fluid, then the pressure upon 
the point B will De due to the pressure of a column of fluid 
whose height is B A. Join A and H, and from any point, 
P, draw P D perpendicular to B H, and because P D = P 
H, the pressure upon the point P, will be due to a column 
of fluid whose height is P D, and so on to any other point 
in the side of the vessel. Let us now suppose, that the 
depth of the water, H B, is 8 feet, and the length of the 
side 6 feet, then the whole pressure upon the side will be 
equivalent to the pressure, or weight, of a mass of fluid of 
the form of the wedge, A B H ; area of the triangle, A B H 

= 8 X Y =-32. 

Contents of the wedge — 6 X 32 = 192 eubic feet. 

Pressure = 192 X 62.5 = 12000 lbs. 

It may be observed, that the pounds pressure upon the side, or 
proposed surface of a vessel, is equal to the weight of a column 
of fluid, whose base is the a rea of the surface, and perpendicular 
height, the depth of the centre of gravity, or middle point of that 
surface. 

g 
— - depth of centre of gravity ; 

8X6 = area of surface. 

o 

• •• 8 X 6 X -5- X 62.5 i^ 12000 lbs. as before. 

• Ques. 10. Find the pressure on a flood-gate whose breadtk 
is 8 feet, and depth 6 feet ? 

/» 
Pressure = 6 X 8 X -^- X ^ 2.5 =1:: 9000 lbs. 

Ques, 11. The depth of water pressing against an em- 



PRACTICAL MECHANICS. 109 

bankment is 9 feet ; required the pressure upon each foot 
of length ? 

1 X 9 X -^ X 62A:^ 253Ulbs. 

Ques. 12. Compare the pressure upon the sides of a cubi- 
cal vessel filled with water, with the pressure upon the bot- 
tom, allowing that the side of the base ^= a foot ? 

Pressure on the base =z a X a X (i X 62.5 ; 
Pressure on the sides =:= ^ x «- X ^ X 62.5 X 4 ; 

Pressure on the sides ] o -n. ^ • xi .r, 

_ • • Pressure on the base [ = ^' ^^^^ ''' ^^'^ P>'^'^"''^ '^'^ ^^^ 
sides is twice that on the base. 

Ques. 13. Required the pressure on the staves of a cylin- 
drical barrel filled with water, the diameter of the base 
being 3 feet, and the perpendicular height 4 feet ? 

Pressure = 3 X 3.1416 X 4 X ^ = 4n2.4 lbs. 

It will be readily seen, that there must be a certain point 
in the side of the vessel when filled with water, where a 
single pressure will exactly counterbalance the pressure of 
the water against the whole side. This point is called the 
ceiitre of pressure. It must obviously lie in the line n G, 
passing through tiie centre of gravity G, of the wedge of 
pressures A H B. 

But, B n == J B H, that is, the centre of pressure 7i, lies 
at one third of B H, from the bottom of the vessel. 

If the staves of a barrel be kept together by a single 
hoop, and if the barrel be filled with water, then the hoop 
must be placed at one third from the bottom. 

Ques. 14. An embankment R D, sustains the pressure of 
water, whose centre of pressure is at P ; it is required to 
determine the conditions of equilibrium, i&c, supposing the 
embankment to turn over upon 0, as a centre ? 

Let V 0, be the vertical line passing through the centre 



110 



PBACXriCAL MECHANICS. 



of gravity of the embankment ; P the centre of pressure 
of the water. Draw P F perpendicular to R 0, then the 
product of the pressure of the water by F, will give the 
moment of the water, tending to turn the embankment over 



R 



A 



H 



P<- 



Q V 



upon as a centre ; and in like manner, the product of 
the weight of the embankment by V, will give the mo- 
ment of the embankment, tending to turn itself in a direc- 
tion opposed to the pressure of the water. When these 
moments are equal, the embankment is upon the point of 
overturning. If the moment of the water be greater than 
that of the embankment, the structure will fall, and vice 
versa. In the following examples, the length of the em- 
bankment is taken at one foot, because, if it stand for one 
foot, it will stand for any other length. 

Ques, 15. Let R =^ 9 feet, D = 3 feet, and the weight 
of a cubic foot of the material 150 lbs. ; will the embank- 
ment stand or fall when the water is up to the top ? 

Surface upon which the water presses :=: 9 X 1 . 

9 
Pressure of the water = 9 X 1 X "5- X 62.5 -- 253U lbs. 

Distance of the centre of pressure from the bottom 

= P = 'I 1= 3 feet ; 

o 

Moment of the water 

= 253U X 3 -^:= 1593.75. 



PEACTICAL MECHANICS. HI 

Wei<-ht of the embankment = 3 X 1 X 9 X 150 = 4050 lbs. 



V — — =1.5 feet. 
^ 2 



Moment of the embankment 

= 4050 X -^ == 6015. 

So that the moment of the water is greater than that of the 
embankment ; hence the structure will fall. 

It will be found when the height is 12 feet, the thickness 5 teet, 
and the weight of a cubic foot of the material 120 lbs., the em- 
bankment will be just upon the point of overturning 

Ques. 16. What must be the height of the water in the 
last example, so that the embankment may be upon the 
point of overturning ? 

Let X be the height, then the moment of the water 

= :.X1X^X62.5X^= ^ X 62.5 ; 

... 4!- X 62.5=6075; 

.\ X zz:^ 8.3 feet. 

Ques, 17. Required the base, or thickness of a rectangu- 
lar embankment, when the height is 15 feet, the Aveight of 
a cubic foot of the material 140 lbs., and the water stands 
at the brim, so that the structure may be upon the point of 
overturning ? 

Pressure of the water 

rr^ 1 X 15 X ^ X 62.5 :::== t031i Ibs. 

Moment = ^ X 103U = 35156.25. 

Let X be the reqmred thickness, then the moment of the em- 
bankment 

^ 15 X r^ X 1 X 140 X 4- ^" ^' X ^^^^ 5 



112 PRACTICAL MECHANICS. 

.*. x' X 1050 = 35156.25 
.•. X -^ 5.79 feet. 

Ques, 18. Let the embankment liave the form of a tra- 
pezoid, A H R C, where A B = 3 feet ; B C = 2 feet ; R 
C = 9 feet, and the weight of a cubic foot of the material 
= 100 lbs. ; will the embankment stand or fall, when the 
water is at the brim ? 

Let the embankment be divided into two parts, namely, the rect- 
angular R B, and the triangular part A B H. Now, the vertical 
line passing through the centre of gravity of the triangle, will cut 




the base at n, 2 feet from the centre of motion A ; and a line 
through the centre of gravity of the parallelogram H C, will cut 
A C at the distance of 4 feet from A . 

Weight of A B H = ^ ^ ^ x 1 X 100 := 1350 lbs. 

Moment of A B H = 1350 X 2 :zz= 2700 ; 

Weight ofBHRC=2X9XlX 100= 1800 lbs. 

Moment of B H R C = 1800 X 4 = 7200 ; 

Consequently, the moment of the whole embankment 
= 2700 -f- 7200 = 9900. 

The moment of the water 

-^ X 62i X -^ 
henc&it follows that the embankment will stand. 



= 9 X 1 X 



= 15931 



PRACTICAL MECHANICS. 



113 



Ques. 19. Required the modulus of stability of the struc- 
ture, R H D, (fio-. example U,) when D = 5 feet ; D H 
= 9 feet ; the weight of a cubic foot of the material in O 
R H D = 150 lbs. ; the water is up to the top ? 

The weight of the structure 

— 5 X 1 X 9 X 150 = 6750 lbs. 
The pressure of the water -= 253 1^ lbs. 

Draw the rectangle R D, take D P = 3, draw P F par- 
allel to D, intersecting P P in* the point C ; from any 
diaa-onal scale mark off C n = 6750 and C t = 2531 ; con- 
struct the parallelogram nt, — then the diagonal Ce pro- 
duced, will intersect the base in Q, the ratio ot Q to U 
V will be found to be about 13 to 25. 

Ques 20. The breadth of a flood-gate is 10 feet, and the 
depth 6 feet, the hinges are placed at 1 foot from the respec- 
tive extremities of the gate. It is required to find the 
pressure upon the lower hinge ? 

In this example, the pressure of the water on 
each half of the gate MC 



— 6 X 5 X 



X 62.5 = 5625 lbs. 



P ^ ^t<t«K 






Let C D be the height of the gate, A and B 
the hinges, and P the centre of pressure of the 
water ; then DP = f = 2;AP=.5 — 2 = 
3 ; and B A r=: 6 — 2 ==:=: 4. 

Now, since the pressure of the water at P is 
supported by the hinges A and B, then by the 
I^iucipal of the lever, sapposing A the fulcrum. 

Pressure on B X 4 = pressure on P X 3. 
Call X the pressure on B. 
... ix = 5625 X 3 
.-. 2:— 4218 lbs. 

Ques. 21. Let H D = 12 feet, D = 1.5 feet, ^veight of 
a cubic foot of the material = 130 lbs., S Y = 5 leet, a 



114 



PRAQTTCAL MECHANICS. 



sboro, or stay supporting the embankment ; S = 3 feet ; 
required the thrust upon the stay, when the embankment is 
upon the point of overturning on 0, the water being at the 
top? 




I) 



When the embankment is supported by a shore, or stay, 
S Y ; and it is required to find the thrust upon this stay 
when the embankment is upon the point of overturning on 
the edo;e 0. 



Weight of the embankmeut 

= 1.5 X 1 X 1-2 X 130 
Moment of the embankment 

1.5 



2340 lbs. 



2340 X 



= 1755. 



Let T be perpendicular upon the stay, then T :::-- 2.4 feet. 
Put z = the thrust on the stay, then the moment of the thrust 
on the stay 

— 2r X 2.4 ; 



The moment of the pressures of the water 

= 12 X 1 X -^ 62.5 X -^ = 18000. 



^ C. X ^ 



;^ X 2.4 + 1755= 18000. 
.-. z -^ 6768 lbs. 



PRACTICAL MECHANICS. 115 

To determine the form of maximum, or equable strength of 
an embankment. 

The embankment, V H N F, will have a maximum 
streno-th, if the part V U, be upon the point of overturn- 
iu<y upon the edge Q, at the same time that the portion V 
h'n F is upon "the point of overturning upon F, by the 
pressure of the fluid. 

Ques. 22. The height of the embankment, H N = 6 feet, 
Y Q ^ 3 feet, and the weight of a cubic foot of the mate- 
j,jal = 120 lbs. ; it is required to find the base F N, so that 
the embankment may be upon the point of overturning on 
F, by the pressure of the fluid, at the same time that the 
part V U is upon the point of overturning on Q, when the 
water stands at the brim ? 




The base Q U, must be first found, so that U V may be 
upon the point of overturning on Q, by the pressure of the 
water on H U. 

Put 2; = Q U, then since the moments of the water and of the 
wall are 

I=Z2:X3X1X120X-|-=1X3X -^-X 62.5 X y ; 

.-. 180x^ = 281.25 
.-. X = 1.2514 feet. 
To find F N, put F D = ?/, then, the moment of F Q 

^^ X 1 X 3 X 120 X -|- ; 



116 PRACTICAL MECHANICS. 

Moment of V D N H 

y X -^ j ; 

The moment of the water 

-^ 6 X 1 X -^ X 62.5 X -3- = 2250. 

Now the sum of the two former moments must be equal to the 
moment of the water, hence by reduction 

180 3/^ + 900 2/ -|- 562.5 = 2250. 

Solving this equation, it will be found that y = 1.45, and F N 

= 1.45 -f 1.25 = 2.t feet. 



EEVETMENT WALLS. 



A WALL sustaining the pressure of earth, or any loose 
material, is called a revetment wall. The pressure of earth 
upon a wall is similar to the pressure of water, with only 
this difference, that the weig'ht of the material must be 
reduced by a certain ratio depending upon the angle of its 
natural slope. The thrust of earth upon a wall is occa- 
sioned by a certain portion, of a wedge shape, tending to 
break away from the general mass. Coulomb showed, that 
the angle which this line of rupture makes with the verti- 
cal, is one half the angle which the 
line of natural slope makes with the 
vertical. Now when the earth is 
level with the top, the pressure of 
earth may always be found by ic- 
garding it as a fluid, having the 
weight a cubic foot, equal to the 
weight of a cubic foot of the earth, 
multiplied by the square of the tangent of half the angle 
of the natural slope from the vertical. In earth of mean 
quality, the angle of natural slope is about 45^, then the 
square of the tangent of half this angle will be == .1716, 
this number, therefore, multiplied by the weight of a cubic 




PRACTICAL MECHANICS. 117 

of the material, gives the weight, in this case, of a cubic 
foot of the equivalent fluid pressing against the wall. 

Ques. 23. A revetment wall 40 feet high, and 10 feet 
thick, sustains the pressure of earth of mean quality, hav- 
ing the weight of a cubic foot equal to 100 lbs. ; it is re- 
quired to determine whether or not the wall will stand, 
taking the weight of a cubic foot to be 120 lbs. ? 

Weight of a cubic foot of equivalent fluid, acting at the back 
Hke water 

= 100 X .ni6 ^ n.l6 lbs. 

Pressure of the earth 

40 
= 40 X 1 X — - X n.l6 = 13728 lbs. ; 

Moment of the earth 

40 

= 13728 X -5- = 183040 ; 

o 

Moment of the wall 

= 1 X 40 X 10 X 120 X -7- ^ 240000 ; 

r(M!S(>nueiit.]y tlie wall will stand 

Ques. 24. What is the thickness of the wall, in the last 
example, so that it may be upon the point of overturning ? 

Let X = the required thickness, then the moment of the wall 

= 40 X 1 X :?: X 120 X "4- — 2400 x'' ; 

when the wall is upon the point of overturning, the moment of the 
wall, must be equal to the moment of the pressure of earth. 

.-. 2400 x' ^ 183040 

,•, X =-= 8.7 feet. 

Ques, 25. Let R = 9 feet, D = 3 feet, the weight 
of a cal)ic foot of the material in R D = 150 lbs., the water 
is to the top H ; besides, the emljankment is supported by 



118 



PRACTICAL MECHANICS. 



R II 




earth L 0, of mean quality, L == 3 feet, and level at the 
top ; will the embankment stand, when the weight of a 
cubic foot of the earth = 120 lbs. ? 

It has also been found by experiment, that when a wall 
is supported by the pressure of earth, the weight of the 
equivalent fluid is about 6 times the weight cf the earth. 

Consquently, the earth L 0, may be considered a fluid, the weight 
of a cubic foot of which = 6 X 120 = t20 lbs. 

.•. The moment of this earth 

= 3XlX-^X720X^= 3240 ; 
Moment of the wall OH 

= 3xlX9xl50X-|-== 60t5 ; 
Moment of the water 

zzz: 9 X 1 X -4- X 62.5 X -|- = ^^93.75 ; 

.-. 3240 + 6075 is greater than 7593.75, 
and hence tbe wall R H D will stand. 



In these examples, the earth is supposed to be level at 
the top, but when the earth has its natural slope, a similar 
method of calculation will apply. When the wall resists 
the pressure of the earth, take \ of the weight of the earth, 
for the weight of the equivalent fluid ; and on the contrary, 



PRACTICAL MECHANICS. 



119 



Avheii the earth resists the pressure of the wall, take i 
the weight of the earth for the weight of the equivalent 
fluid. 

Ques. 26. Required the same as in example 23, when the 
earth has its natural slope Q R, and the thickness of the 
wall = 8 feet = B C ? 



tl Q 




B C 



On this supposition, the weight of a cubic foot of the equivalent 
fluid 

== 100 X i = 12i lbs. 

The moment of the earth C Q R 

40 40 

= 40 X 1 X -y- X 12i X -^ == 133333 ; 



The moment of the wall 

= 40 X 8 X 1 X 120 X 
Consequently, the wall will stand. 



153600 ; 



Ques. 27. A revetment wall, F Q B C, is 30 feet high = 
B F ; B C =6 feet. On one side, G Q, earth of mean 
quality is sustained level witli the top, and on the other side, 
F B, the earth has a natural slope A D, and rises to the 
height B D =^ 5 feet ; will the wall stand or fall, supposing 
that the weight of a cubic foot of the earth to be 120 lbs., 
and that of the wall 130 lbs. ? 



^-•^ PRACTICAL MECHANICS. 

levd Of 'tll'rw^l?' "^•^™^"* ?! f't e^rth, Which rises to the 
le\el ol the wall, is opposed bv the sum of the moments of 
the^wall, and the earth on the other side having rnatural 

Weight of the wall 

= 30 X 1 X 6 X 130 = 23400 lbs. 
Moment of the wall 

= 23400 X -|- == 70200 ; 

theTloptgtar'tir'" '^^^ '' ''' ^^"'^^'^"^^ ^^''' ^^^ -^Pect to 

= 120 X i = 60 lbs. 
Pressure of this earth 

= 5 X 1 X 60 X "4- = 750 lbs. 

Moment = 750 x — = 1250 ; 
Total moments sustaining the wall 

r= 70200 + 1250 = 71450 
Moment of the level earth 

= 30XlX^Xl20x.ni6X^ = 92664; 

Since the latter moment is greater than the sum of two the 
former, it may be concluded that the structure will faU. 



OP FLOATING BODIES, AND SPECIFIC GRAVITY. 

.nrJof "Ilf/fl*^-^! ^^^*',' ^* i^«»stained by the upward pres- 
« re, t^l 'V ^""^ ^' *'''^'*' ^^ ^" equilibrium of pres- 

uies, the upward pressure must be equal to the <rravity of 
the floating body. But this upward pressure would last 
support a volume of fluid equal to t!,atSvhich is d1 placed 
therefore the weight of the floating bodv must be em al to 
the weight of the fluid displaced. ^ ^ 



PRACTICAL MECHANICS. 



121 



Hence it follows, that if a heayy body be weighed in 
water, the weight which is lost will be equal to the weight 
of water having the same bulk or volume as the body. In 
this way we are enabled to find the specific gravity of a 
body, or its weight, as compared with the weight of an 
equal bulk of water. As the weight of a cubic foot of 
water is just 1000 ounces, it is customary to consider the 
specific gravity of a body as the weight of a cubic foot of 
it. The following table contains the weight of a cubic 
foot of various kinds of material in construction. 



NAME OF 
THE MATERIAL. 



k 2 2 ai 

2 S B o 

^^ < ^. 

w 5? fe 

, ^ o § 



Ash, 

Beech, 

Deal, 

Deal, Memel. . 
Fir, American, 
Fir, Riga. . . . 



760 
690 
698 
590 
553 
753 



NAME OF 
THE MATERIAL 



Brass, cast. . . 
Iron, wrought 
Iron, cast. . . . 

Lead, cast 

Oak, English.. 
Teak 



>• c- O 

H a o 
^ S ^ aj 

« Is S 

C3 !=^ S S". 
a ti t) 

o £ c^ O 

B^ ^^ 



8399 

7700 

7066 

11332 

934 

657 



Ques. 28. A barge, (supposed for the sake of simplicity, 
to be of a rectangular shape), is 10 feet long, 5 feet broad, 
and 4 feet deep, outside measure, the thickness of the plank- 
ing is 2 inches, and the weight of a cubic foot of the tim- 
ber is 50 lbs., to what depth will the barge sink when 
loaded with 4 tons 1 

Content of the exterior s«lid 

= 10 X 5 X 4 = 200 cubic feet. 
Content of the interior solid 

= 91 X 4f X 3f = 172.92. 
200 
172.92 



Timber in the barge, 27.08 cubic feet. 

Weight of timber in the barge 

= 27.08 X 50 = 1353 lbs. 



122 PRAfTICAL MECHANICS. 

Let X be the depth of the water displaced, then the weight of 
the displaced water 

= a: X 10 X 5 X 62.5 = 3125 x; 

.\ 3125:?:= 1353 -f 4 X 2240 

,•, 2; ,= 3.3 feet. 

Ques, 29. How many tons will just sink the barge in tlie 
last example ? 

Let 3/ = the number of tons ; 

200 X 62.5 = 12500 ; 

.•. 12500 = 1353 + y X 2240 

.•. y i:^ 4.9t6 tons. 

Ques. 30. What weight will sink the barge in example 
28, to the depth of 3 feet ? 

10 X 5 X 3 X 62.5 = 9375. 
Let the required load = z^ then 

9375 ^ 1353 + z X 2240 
,*. 2r := 3.58 tons. 

Ques. 31. How far will the barge sink when empty ? 
Put V = the required depth, then 

10 X 5 X ^ X 62.5 = 3125 v; 
.'. 3125^=1353 
V = .43296 feet. 

Ques. 32. A body weighs 49 grains in air, and 42 grains 
in water ; what is the weight of a cubic foot of the 'sub- 
stance ? 

The loss of weight 

= 49 — 42 = 7. 

This loss is the weight of water having the same bulk as the 
body. 



PRACTICAL MECHANICS. 123 

Consequently, the number of -times the body is heavier than 

water 

49 
= — -— ^^ 7 times ; • 

But the weight of a cubic foot of water = 1000 ounces, there- 
fore, the weight of one cubic foot of the body = 1 times 1000 
ounces = 7000 ounces, and this number is called the specific grav- 
ity of the body. 

Ques, 33. A cubic foot of timber sinks 9 inches in water ; 
required the specific gravity ? 

In this case, the weight of the displaced fluid will be the weight 
of a cubic foot of the timber. 

Therefore, the specific gravity 

= 1 X 1 X ^ X 1000 = 750. 

Ques. 34. A Solid, whose weight is 60 grains, weighs 40 
grains in water, and 30 grains in sulphuric acid ; required 
the specific gravity of the acid ? 

Weight lost in water = 60 — 40 = 20 grains • 
Weight lost in acid = 60 — 80 = 80 grains. 
Therefore, the number of times the acid is heavier than water 

— 20 — '^ ' 
Hence, the specific gravity of the acid 
= li X 1000 = 1500. 

Ques. 35. A piece of metal weighing 36 lbs. in air, and 
32 ^ lbs. in water, is attached to a piece of wood whose 
weight is 30 lbs., and then the compound mass is found to 
weigh 12 lbs. in water ; required the specific gravity of the 
M'ood ? 

Weight of water, equal in bulk to the metal 
= 36 — 82 _:. 4 lbs. 



124 PRACTICAL MECHANICS. 

Weight of water, equal in bulk to the compound 

= 36 -f 30 — 12 = 54 lbs. 

Weight of water equal in bulk to the wood 

= 54 — 4 = 50 lbs. 

But the weight of the wood is 30 lbs., that is bulk for bulk, the 
wood will be three-fifths of the weight of the water. Conse- 
quently, the specific gravity of the wood 

3 
= -^ X 1000 = 600. 



OF THE LIMITING ANGLE OF RESISTANCE. 

Let W be a material particle acted upon by a pressure 
P, in the direction of and equal in magnitude to P W. 
Complete the rectangle A C, then will C W be the pres- 




sure upon the plane, and A. W that part of the force P, 
which tends to give motion to the particle along the plane. 

Put a for the angle, P W C, then. 

Pressure on the plane = C W — P cos. a ; and the resistance 
of friction 

=: f P COS. a ; 

The efi*ective pressure tending to move the particle in opposition 
to friction 

— A W = P sin. a ; 

Motion will or will not take place, according as their 
pressure is greater or less than the resistance of friction ; 



PRACTICAL MECHANICS. 125 

and when motion is upon the point of taking place, the one 
must be equal to the other, and then the angle a, is called 
the limiting angle of resistance ; 

Therefore in this case 

/ P COS. a ^=:^ P sin. a 

sin. a 

.•. / =■ = tan. a, 

COS. a 

This equation shows that the co-eflficient of friction is 
equal to the tangent of the limiting angle of resistance. If 
the pressure be applied within the angle, then no motion 
can take place, however great that pressure may be ; and 
on the contrary, if the pressure be applied without this 
angle, then motion will take place, however small that pres- 
sure may be. Foreign writers make this property the basis 
of their theory of machines. 

Ques. 36. The co-efficient of friction, / = .683, what is 
the limiting angle of resistance ? 

From a table of natural tangents it will be found, that the angle 
34°. .20' has a natural tangent nearly equal .683 ; hence, the limit- 
ing angle of resistance := 34°..20^ 

Ques, 37. Equilibrium of the lever, and the wheel and axle, 
taking the friction upon the axis into account ? 

Let P Q be a lever, having the circular axis C T, turn- 
ins: within a circular socket. In order that motion should 




take place in the direction of the pressure P, the resultant 
of the two pressures P and Q must fall on tlie left side of 
the centre of the axis, and by the last problem, this re- 



126 PRACTICAL MECHANICS. 

sultant C T, must be inclined to T at the limiting angle 
of resistance. 

Taking T, therefore, as the centre of motion, by the equal- 
ity of moments for the state bordering on motion, 

PXPF = QXQF. 

Ques. 38. Let P = 14 inches ; Q =- 12 inches, P = 
27 lbs., and angle T F = 30", T = 3 inches ? 

The natural sine of 30° = -.500000.* 

OF =sin. a 30° X 3 = 1.5 ; 
PP = 14 _ 1.5 = 12.5 ; Q F = 12 + 1.5 = 13.5 ; 

.-. 2t X 12.5 = Q X 13.5 
.•. Q = 25 lbs. 

• Ques. 39. Required the same as in the last example, when 
the angle T F = 40°, and P = 100 lbs. ? 

Natural sine of 40° ^ .642^876 ; 

,643 will be near enough for practical purposes. 

P =:= sin. 40 X 3 = .643 X 3 = 1.929 ; 

PP:^ 14_ 1.929 = 12.0n ; QF = 12+ 1.929= 13.929; 

.•. 100 X 12.071 = Q X 13.929 ; 

.•. Q ^ 86.6 lbs. 

Ques. 40. Let P be the radius of a wheel, and Q the 
radius of an axle ; put W for the weight of the wheel and 
axle, then as this weight acts through the centre 0, then, 

PXPF = QXQF + WX0F; 

Suppose the radius of the = 20 inches, the axle 4 inclies, 
and the axis 2 inches ; required Q, when the power applied 
to the wheel = 100 lbs., the weight of the wheel and axle 
=: 80 lbs., and the limiting angle of resistance = 21°. 



* See the " Practical Model Calculator J"* by Oliver Byrne, the author 
of the present work. 



PRACTICAL MECHANICS. 127 

The natural sine of 2P :=:= .3583679 

.358 is near enough in practice. 

OF — sin. 21° X 2 = .716 ; 

p F z= 20 — .716 = 19.284 ; F = 4.716 ; 

P = 100 ; W= 80 ; 

.•. 100 X 19.284 = Q X 4.716 + 80 X .716 

.•. Q = 396.7 lbs. 

Ques, 41. Let the pressures P and Q be inclined to tlie 
lever, then produce them till they intersect at C, and join 
the points C and ; now the resultant of the forces P and 
Q must be drawn, so that the angle T C = the limiting 
angle of resistance ; for this purpose, describe a circle upon 
the chord C 0, so as to contain the limiting angle of resist- 
ance ? 




Let this circle intersect the circumference of the axis at 
the point T, join T and C ; then on the line C P take off 
a space n, equal to the units in the pressure P, and con- 
struct the parallelogram of pressures C n m t, then the 
units in C ^ will give the pressure of Q, when the motion is 
upon the point of taking place. 



128 



PRACTICAL MECHANICS. 



Values of/, or tan. a, according to the experiments of M. 
Morin. 

Iron on oak 62 

Cast iron on oak 49 

Oak on oak, fibres parallel , , , , .43 

Oak on oak, greased 10 

Cast iron on cast iron ....,, ,15 

Wrought iron on wrought iron 15 

Brass on iron 16 

Brass on brass 20 

Wrought iron on cast iron 19 

Cast iron on elm 19 

Soft limestone on the same , .64 

Hard limestone on the same. 38 

Leather belts on wooden pulleys 4^ 

Cast iron on cast iron, greased 10 

Brass on iron, greased 08 

Pivots, or Axis of wrought iron, or cast iron^ 

or brass, or cast, iron pillow blocks ; 

Constantly supplied with oil , , .05 

Greased from time to time 08 

Without any application, or dry 15 



TO DETERMINE GENERALLY THE FORCE OP TRACTION. 

Let W be a heavy body drawn alone the horizontal plane 
A W, by the force of the traction P, acting in the direc- 




tion and with the magnitude W P. This force into W A, 
and W C, the former force just overcomes the resistance 



PRACTICAL MECHANICS. 129 

of friction, the latter tends to reduce the pressure of the 
body on the plane ; hence, |putting b = angle P W A. 

Pressure of W on the plane = W — P sin. b. 

.•. the resistance of friction 

= f(W — F sin. b.) 

But when motion is about to take place, this resistance is equal 
to the force W A, 

or = P COS. b ; 

.•• P COS. b = f W — sin. b) ; 

. r ^(^' • (z) 

COS. b + J sin. b ^ 

In the same manner, the traction may be found when the body 
is moved upon an inclined plane. 

Ques. 42. A stone weighing 2 cwt., is drawn along a 
horizontal plane, by a cord inclined at an angle of 35° ; 
required the force necessary to move the stone, supposing 
the co-efficient of friction to be = .70 ? 



Natural sine of 35° = 


.809 } '^^^'•^y- 


Natural cosine of 35° = 


.T X 2. 


— ^ = 1.15 cwts. 



.809+ .1 X .574 

TO FIND THE LEAST TRACTION. 

In equation (Z), substitute the value of/ before given 

sin. a 



COS. a 
tlien by an easy reduction, it will be found that, 
W sin. a 



COS. (b — a) 



It is evident that P will be the least possible, when the 
denominator of this expression is the greatest possible, 
which will obviously happen when 



130 PRACTICAL MECHANICS. 

h — a ^= Oy 
ov b = a ; 

that is, when the angle of traction is equal to the limiting 
angle of resistance. 

Therefore, in this case, 

P = W si7i. a; (J). 

Ques. 43. A stone weighs 10 cwt., and is drawn along a 
horizontal plane ; required the least traction, taking a = 
23°; 

By equation (J) ; Nat. sin. 23° — .3907 

P r= 10 X .3907 = 3.907 cwts. 

Ques. 44. Required the least traction, when the weight 
Df the mass is 9 cwt., and the rubbing surfaces are soft 
calcareous stone upon the same, and the limiting angle of 
resistance = 36°..30'. 



JSTatural sine of 36°..30' 

Least traction 1= 5.3532 cwt. 



.5948 
9 



Ques. 45. If a heavy body be moved upon a rubbing sur- 
lace, lying between two given points, by a pressure acting 
parallel to the surface, the work is always the same, what- 
ever may be the form, of the surface. 




Let W be a body moved on the plane A C, then by the 
resolution of pressures, the pressure of W on the plane = 
W COS. A ; W qmn, being the parallelogram of pressures, 
and since W n is perpendicular to A C, and mn parallel 



PRACTICAL MECHANICS. 131 

to it, the angle m W n = the angle A ; therefore, the re- 
sistance of friction 

— f W COS. A ; 
Consequently, the total work on the plane A C 

z:= / Wcos. AxAC-{-WxCB 
-_^fWxAB-\- W X C B. 

This final expression is independent of the inclination of 
the plane, it being, in fact, the work expended in moving 
the body over the horizontal distance A B, added to the 
work due to gravity in elevating the body to the vertical 

space B C. i . ^ 

As a curved surface may be regarded as being made up 
of an infinite number of straight planes, therefore, the work 
upon the whole curve will be equal to the work done upon 
the horizontal projection of the curve, added to tlie work 
done -in opposition to gravity. When bodies descend a 
f=urface, the work of gravity becomes minus. 

This total work is obviously, iudependent of the nature 
of the curve. In general, it may be shown, that whatever 
may be the zigzag course pursued by the body, the work 
will always be equal to the work on the projection of this 
curve, added to the work due to gravity. When the in- 
clination of the plane is small, the horizontal distance may 
be taken equal to the length of the plane. See Chapters 
1. and II. 

Ques. 46. What work would a horse be able to perform 
in drawing a load of one ton to the horizontal distance 
of two mites, up a curved incline, whose total elevation =-= 
500 feet, the co-efificient of friction being ^- 3V ? 

Work on the horizontal line 

^ _:^M®_ ^^2.x 5*180 = -iS«*AOO ; 

Woik due lo gravity 

x= '1^4« X 5i>« = IViOOOO, 
Total work = =53*J*£00 + 11*£000« = 1859*AOO, 



132 



PRACTICAL MECHANICS. 



If the body W, be upon the point of descending the plane 
by the force of gravity alone, then, the work necessary to 
. move the body is nothing ; hence, 



fWxAB— WxCB 
• •# / = tan, A ; 
a result before obtained. 



O 



THE PARALLELOGRAM OF PRESSURES PROVED ON THE 
PRINCIPLE OF WORK. 

Let B D, and A C, be two inclined planes intersecting 
each other at right angles in the point W. Let a body W, 
be placed between the two planes ; then the force with 




wnich the body tends to descend the plane D B, will be 
equal to the pressure which it exerts upon the plane A ; 
and the tendency to descend the plane C A will be equal 
to the pressure exerted on the plane D B. 

Let the vertical line W H be drawn through the body, 
and take W H = the units of weight in W ; from H let 
fall H C, perpendicular to the plane A C, and H D perpen- 
dicular to the plane B D, then putting A and B for the in- 
clinations of the planes A C and B D, by trigonometry 

W G -^ w X sin, A = 

the tendency down the plane A C. 



PRACTICAL MECHANICS. 



133 



Similarly, W J) = w sin. B = the tendency down the plane 
B D. But these forces W C and W D, are generated by the 
gravity of the body, or a force represented by the diagonal 
W H. Having thus established the proposition for the 
case of the rectangle, it may be very readily extended to 
the general form of the parallelogram. 



THE EQUILIBRIUM OF THE ARCH. 

When the centres of an arc are taken away, the crown 
almost invariably sinks ; this occasions the joint at the 
crown to open at its lower edge, and at the same time a 
certain portion, D V P N, of the arc to turn upon D as a 
centre, thereby producing a rupture, or opening, in the ex- 
terior edge at this point. The same effect will take place 
in the other half of the arch. 




These two equal portions which thus tend to break away 
from the general mass, exert a horizontal pressure along 
the line P C, thereby tending to cause the walls of the 
structure to turn on their outer edges. The arc will under- 
go a rupture at that point where the portion D A^ P N, so 
breaking away, will produce the greatest horizontal thrust ; 



134 PRACTICAL MECHANICS. 

for this point must, obviously, be the yielding point of the 
arc. 

To find the point of Rupture. 

Let B Gr, be a* vertical line, passing through the centre 
of gravity of the mass D V P N, and intersecting P C in 
the point B ; join the points B and D, take B equal to 
the units of weight in the mass D Y P N, and construct the 
rectangle F C B, then the units in B C will be the hori- 
zontal thrust, if it be greater than any other so determined, 
and the point D will be the point of rupture. MM. 
Clapeyron and Lam^e, found that the resultant D B forms a 
tangent to the intrados of the arch, when the line of rup- 
ture is assumed to follow the vertical line D V. This prop- 
erty gives a very easy method for finding the point of 
rupture ; for if upon constructing the figure, as already 
described, it is found that B D touches the curve, then I) 
will be the point of rupture. The most troublesome part 
of this calculation, consists in finding the centre of gravity 
of the mass D V P N. However, the following method 
will be found sujfficiently accurate for practical purposes : 

If D Q be divided into n equal parts, ai*d lines be drawn 
from the points of division perpendicular to D Q, cutting 
the intrados and extrados ; let x and y be put for D Y and 
N P, the extreme ordinates respectively, and A, B, C, &c., 
for the first, second, third, &c., intermediate ordinates, or 
vertical distances between the intrados and extrados, then 
putting s = the common distance between the ordinates 

six + (3 7?.— l)Y + 6(A-f2B -f3C + &c.) j 

■pv n __-i 

3|X+Y+2(A + B+C + &c.) I 

If that part of the divisor which is within the brackets 
be multipled by | it will give the area of D Y P N. 

Having ascertained the horizontal thrust, B C, the centre 
of gravity of the whole mass, including the semi-arch X P 
N H, with its load, if any, and the pier H X ; then suppos- 
ing the pier to turn upon its outer edge, if the moment of 
this mass exceeds the moment of the horizontal thrust, the 
structure will stand, and vice versa. 



PRACTICAL MECHANICS. 



i^i 



Ques. 47. The radius of a semicircular arc is 11 feet, the 
thickness of the crown 18 inches ; the mason work is built 
level with the crown, and the weight of a cubic foot of the 
material = 120 lbs. ; required the point of rupture, and 
the horizontal thrust, taking a depth of 1 foot of the 
masonry ? 




Construct a figure on a scale of an inch to the foot, and 
dividing D Q into six equal parts, it may be found by calcu- 
lation or measurement, that X = 7.8, A = 5.4, B rr= 3.8, 
C = 2.8, D = 2, E = 1.6, Y = 1.5, and n = 6, and D Q 
= 10, and hence s = V^« 

By the formula it will be found, that D G r=rr 3.53 feet, 
and the weight D V P N = 4000 lbs. Take B == 4000, 
and construct the parallelogram 0, then B C = 1900 lbs. 
nearly ; and as B D forms a tangent to the curve, the point 
D will be the point of rupture. The point D is 65"^ from 
N, the crown. 

Ques. 48. If the piers in the last example be 4 feet thick, 
and their height 28.5 feet measured to the level of the 
crown ; will they stand or fall ? 



136 PRACTICAL MECHANICS. 

In this case, tne weight of the whole semi-arch ^^ 5100 lbs., 
taking as before the depth, one foot. 

The distance of the centre of gravity from the outer edge of 
the pier =7.6 feet. Weight of the whole pier :=i: 4 X 1 X 
28.5 X 120 = 13680 lbs. 

Hence the moment of the pier 

= 13680 X 2 = 21360 ; 

The moment of the semi-arch 

= 5100 X 16-^38160 ; 

and the moment of the horizontal thrust 

^ 1900 X 28.5 = 54150 ; 

The sum of the two first moments being greater than 
fche moment of the thrust, it follows that the pier will 
stand. But it will be found in the same manner, if the 
pier be only three feet thick the structure will fall. When 
the pier is 3.3 feet thick it will be on the point of over- 
turning. 



CHAPTER Vni. 

WOODEN AND IRON BRIDGES. 



The equilibrium of wooden and iron bridges will be 
considered after I give some investigations, illustrating a 
higher method of investigating mechanical subjects. 

Variable motion is caused by a force, which, acting con- 
tinually upon a moving body, goes on at every instant in- 
creasing or diminishing its velocity. Such force is measured 
at each instant, by the ratio of that very small increment 
or decrement of velocity, w^hich it causes in the moving- 
body, to that very small time in which the increment or 
decrement is caused. When this ratio is constant, or so 
that in equal times equal degrees of velocity are added to, 
or taken from the moving body, the accelerating or retarding 



PRACTICAL MECHANICS. 137 

force is constant, but when this ratio is variable, the force 
is also variable. 

It is more convenient to use the symbol [~ instead of the 
usual symbol d, employed in the calculus ; [ "~ must not be 
taken for ^, which is the well-known sign of the square 
root. 

Instead of dy, dx, &c., I write 

I y, I X, &c. 

In the case of a variable motion, Put P for the acceler- 
ating force, Fthe velocity, S the space, and T the time, the 
relations between these qualities may be found by the equa- 
tions 

V^ ^- (I) • 

Other general forms may be deduced from these, for ex- 
ample, from (I), 

re" 
[T = 4^, and from (II), 

— •, and 



V F 

.-. F[S =^V (F'. 

Again eliminating V from (I) and (II), 



Ffr 



m. 



On the supposition that T is the independent variable, the 
last equation becomes 



188 PRACTICAL MECHANICS. 

2 



The above equations relate to accelerated motions, by 
making |~F negative, they are adopted to retarded motions. 

In an equally accelerated motion F, is constant, from (TI) there- 
fore, 

f F\f --^ f\v, integrated 

becomes F T = V; (III), 

The velocities, therefore increase as the time when the 
force is constant. It is not necessary to correct (HI), when 
the body moves from a state of rest, for then when T = o, 
V =^ 0, and F T == o. But if the body has a velocity a, 
already impressed, then (III) requires correction and be- 
comes 

Y = a + FZ (IV) ; 

for V, becomes a^ in (IV), as it should do, when T^= o. 

When the force is retarding, V becomes negative and (IV) 
becomes 

Y ^a — F T 

It must not be forgotten that \~ stands for the symbol d. 
Again integrating F\s ^=^- T^ | F, 

S = -T-rr + constant. (T). 

2i' ' ^ 

If tlie body moves from a state of rest (V) requires no 
correction, or addition of a constant, for in this case S must 
be equal to 0, when F= 0. But if the body should have a 
velocity «, before the force F commences to act, then the 
space passed over in the time T will be 



2i^ '^F 



7 > 



PRACTICAL MECHANICS. 139 

_ V — a' 

because, when T = 0, the body had a velocity a, and hence must 
have passed over a space 



ft2 
~ 2F 


, before, 


^- 2F 




commences to be described. 





(VI), 



Equation (VI), is readily obtained by integrating and 
correcting the general equation JP [^ = V [ F , the result in 
this case is the equation 



S^ 



a'' — V 
2 F 



Another sent of general expression may be thus formed, — 
From (II), 

p [ T = [1^, integrating and 
F T — V = j=, according to (I), 

.-. {~S ^ F T\T 
integrating, and the result is 

S = -?^^ + constant ; (VII. ^ 

If the body move from a state of rest, (VII) requires no 
correction, for S and T vanish simultaneously, in this case 
also, by transposing 

^ S 



and consequently the measure of the constant accelerating 
force F is obtained, when the space, vrhich the moving body 
has described in a given time T, is knovrn. Near the sur- 
face of the earth gravity may be considered a constant 



140 PRACTICAL MECHANICS. 

force, and since a heavy body falling from rest describes 
I&T2 feet in one second 

F = 32i feet, 

which is generally represented by the letter g, by most 
writers and mechanics. 

It must be remembered that [""" stands for d of the Cal- 
culus, and must not be taken for ^% the sign of the square 
root. 

Correcting (VII) for accelerated and retarded motion, 
with a constant velocity a, before the force F commences 
to act, 

F T" 

S = a T -\ ~ — , (accelerated.) 

F T" 

S^a T — , (retarded.) 

The reasons for these last expressions are obvious, for 
supposing the force F to cease to act during the time T 
then (VII) becomes 

S =:= + constant ^^ a T 
the uniform velocity a multiplied by the time T, 



THE VERTICAL MOTIONS OF HEAVY BODIES THROUGH 
RESISTING MEDIUMS. 

For want of sufiBcient experiments, Mathematicians take 
for granted, that the air and other fluids offer resistances 
proportional to the squre of the velocity of the body mov- 
ing in them. 

Let V be the velocity, then this retarding force may be 
expressed by v '^ multiplied by a constant co-efficient. A, so 
that the retarding force is A v ', the value of A will depend 
on the figure of the body, and upon the ratio of its specific 
gravity to that of the fluid. Consequently, if a body de- 
scend from a state of rest through a resisting medium, the 



PRACTICAL MECHANICS. 141 

accelerating force is g — Kv\ whicli put for F in (I) and 
(II) gives, putting t for the time, and s the space, 

(g — Av'') \~ ~— \T, and 
(g — Av') [S ^ [V. 

For the convemence of calculation put m = the square 
root of g, and n = the square root of a ; then k.=- gn\ 
then the two last equations become 

There will result the two following equations by inte- 
gration and correcting the results on the supposition, that 
when t = Oj s = 0, and v = o ; 

V = ;^^ — . — :r^ and s z:^ ■— —- log -:, ^ — r J 

a third equation may be found by eliminating v, 

1 1 1 ( g'^t —gnt \ 



CURVILINEAR MOTION. 

Suppose the position of the moveable point to be refer- 
red to three coordinate axes, and let the acting forces be 
reduced to three, P, Q, R, parallel to these coordinates. 

Then [T being the small arc which the point will de- 
scribe in the time [T, its velocity at this point will be 

[^ 
[t 

because [T is the diagonal of a parallelepiped, which has 
for its sides [^, [y, [T, the moving point cannot, in the 
instant [7, describe the small arc [T, without at the same 



142 PRACTICAL MECHANICS. 

time describing, in the directions of the ordinates of x, y, z, 
the small spaces ["^, [y, [T; for instead of the velocity 

1^ the point may be imagined to move with the veloci- 
[T ' ties 

[T ' [T ' [T ' 
parallel to the ordinates of x, y, z^ hence the equation 

derived from (I) and (II), gives three following equations, 

Integrating these equations, the partial velocities 

\ X \ y \ z 

are obtained, from the composition of which results the ab- 
solute velocity of the point. Integrating again, and the 
ordinates x^ 3/, z, are found in terms of t, and hence the 
place of the point at any instant. When t is eliminated, 
there will remain two equations between x, y, z, the equa- 
tions of the curve described by the moving point. 

The student must keep in mind that [~" stands for the 
differential symbol d. 

When the forces acting on the moving point are all in 
the same plane, the trajectory will lie wholly in that plane, 
and its equation will be found by eliminating ^, from the 
two equations. 

which may be written 



-= i ^ nnri O — X 



^-. andQ = -i.^, (VIII.) 

t being considered the independent variable, and f"^ is put 



PRACTICAL MECHANICS. 143 

instead of d'' and, [" ^ instead of {dxY, which is written 
d X'. 

Supposing the ordinates to be rectangular, then 



Multiply (VIII) by \^, [7, &c., and add together, 

2 _2_ _ j 

p [T + Q [7 + i^ F = ^^ ^^ + ^^^-[^ + ^ -^^^ 



2 

• 



.-. P [T + Q [7 + J? [T 



2 



.'. P{-7 + Q\J + R\T=-v\l>, (IX.) 
because, _ 

V = 4=^ and 



r — [ « r * 

I ^ 

On ^Ag motion of Cannon Shot. 

If a body be projected obliquely, with a velocity due to 
the altitude A, (the resistance of the air being neglected), 
aud if the abscissa of x be taken in the vertical line drawn 
through the point of projection, and the ordinates of y 
parallel to the direction of projection, the equation of the 
curve described by the projectile will be 

.y' = 4 hx. 

Taking the general equations (VIII,) 



144 PRACTICAL MECHANICS. 

P [T = f -[^ and Q [T =- I" ^^ 



[T - ' ' I [T ' 

putting P — g and Q,-^ o^ and integrating, then 

--^ ^^ g t + constant ; -^~r- == constant. 

To find the constant quantities, it must be observed that 
when t = Oj the velocity in the direction of x = o, 

. • . -L^ + constant = o, hence constant o ; 
\ i _ 

requires no correction. Again, when t = o, the velocity 
in the direction of y is that due to the height A, which is 

constant = (2 gh) ^, the required constant. 



y 



[T 



••. l^ = (2gh)^, (XL) 

Integrate (X) and (XI), and consider t, as before, the in- 
dependent variable, then 

x = igt'siiidy=^ {^g^)^t, 

tliese results require no correction, for when t = o, x = o 
and y = 0,^ condition which the resulting equations satisfy 
without the aid of a constant quantity. Eliminate t, then 

y^^ihx, (XII.) 

an equation to the common parabola, having for its diame- 
ter the vertical drawn through the point of projection ; 
and the parameter is 4 times the height due to the initial 
velocity of the projection. 

Let the coordinates be changed, and instead of taking 
A P for the abscissa of x, and P M for the ordinate of y, 
assume the horizontal line A Q - a:, and the vertical Q M 



PRACTICAL MECHANICS. 

= 3/1 ; let the angle of elevation TAB 
R parallel to A B. 



145 

0^ and draw P 




M R zLz: P M sin. ^; P R = P M cos, 6 ; 
«r a^j + 2; = y sin. ; y^ = y cos. 0, 

. • , Xj :::^ t/, tan. 6 X 



But (XII), X : 



y 



Vi 



x^ = y^ tan. 6 — 



4 h cos. ^ 



3/i 



4 k cos. ^ 6 ' 



(XIII.) 



THE PATH OF A PROJECTILE IN THE AIR. 

Let g r express the resistance of the medium. This force 
being in direction of [T ; now, if the curve be referred to 
an horizontal axis as abscissas of x, and the ordinates y 
vertical, the resistance g r may be resolved into two forces 

\ ^ -, \ y 

\ s \ s 

one parallel to the axis of x and the other to that of y. 
Consequently, the body will be acted upon in the direction 



146 PRACTICAL MECHANICS. 

of X by the force — 9; r —=:- and at the same time in the 

direction of y by the force — g — gr -~~~ . Then the 

I « 
general equations (VIII,) are easily applied, they become 

-gr-\^\T= [4^, (XIV.) 

- -^rT-^riI-[T= T-j^, (XV.) 

From these equations t is to be eliminated. On the sup- 
position that ["^ is constant 

__ 2. 

I » h 

... gr [T' = [T [T, (XVI.) 

_ 2 

In works on the calculus \ t^ is written df", and |"T is 
written d '' t, the symbol d is always represented by [~, and 
is not to be taken for y, the sign of the square root. 



2 



gr \ t 



jC\. 1 


' ±j 11 • 


[^ 


> 




r 


\J _ 


2 

[7 


[7 
[ 


2 

[T 


= 


2 

[7 
" [7 


r7 


gr 


[7 




2 


— gr 




[T 



PRACTICAL MECHANICS. 147 

Consequently, (XV) becomes 

- g \T' = [7, (xvii.) 

From (XVII) another equation is readily found, 

[7 = 2 ^ [T [T, (XVIIL) 

2 

[T, [T, are easily eliminated from- equations (XVI), 
(XVII), (XVIII), hence 

2r f7^ + \T \J = 0, (XIX); 

which is the equation of the trajectory ; when the laws of 
resistance are determined, the curve described by the pro- 
jectile becomes known from (XIX.) 

Suppose the resistance to be proportional to the square 

of the velocity, then ipnt gr = gn' v^ — gn'' — ^ . 

But from (XVII) 



2 



—2 — \y 



g 

. FT' 
.«. r — — gn^ — 2- 

[7 
Hence the equation of the trajectory (XIX), becomes 

3 

Sf.^2gn [T, (XX.) 

■4^ divided by -^ 

2_ ' • \ X \ X 

-^~ , hence, Integrating (XX), 

r7 



It is easily observed that f -4^ divided by -L^, gives 



Loo-. JJi- ::::r. 2 /r ?r S + C 



148 PRACTICAL MECHANICS. 

2 

or,-g, = CE ^Sn's^ ,^^j_^ 

I ^ 

C being the constant required after integrating, and log 

2 

c =^ C. To find c, for [y put its value — g [T\ then 

Now, if be the angle of elevation, and h the height 
due to the velocity of projection, at the beginning of the 
motion, or when 5 = 0, the horizontal velocity 

^"^ -^ {2gh)icos.O. 



.*. c = 



2 h CQS.' • 
• • . (XXI) becomes 

2 A ^2gn's 

In order to abridge, put 2 gn^ ^^ m, and -^-i^ = p, 
then (XXII) becomes T x 

Since IT' + [7= = [T' •• • (1 + p')-' R = R, 
and multiplying (XXIII) by this last equation, 

r7(i + P-)' --fiJ^T, 



2 /I > 



integration will give the following equation. 



^ms 



PRACTICAL MECHANICS. 149 

Putting P for the left hand member of this equation, and 
introducing the value of -p=- from (XXIII), the last equa- 
tion becomes _ 

m I X 

.-. r^ = ^ ^-^ , and 



P m — Cj m 



f s .,- f 



PH 



Pm— C,m'^ •^ Vm—Q,m' 

The constant Ci is found by considering that when s = 
Oj p =^ tan. 0, If these formulas could be integrated, x and 
y would be found in terms of p ; then eliminating _p, the 
equation of the required curve would be found. But since 
the formulas cannot be integrated in finite terms, the fol- 
lowing method of approximation may be employed, when 
the student cannot apply the Theorems of Laplace and 
Lagrange. 



INTEGRATION BY APPROXIMATION. 

Let 2/ = / X ["^ ; if when a: = a, then y = b ; and if 
when 

z =^ a, a -{- d, a -{- 2 d, a -\- ^ d .... a + n d ; 

then X = A, Ai, A 2, A3, ... An. 

and if the difference d be very small, the value of jf X ["x" 

from 

X = a to X ^^ a -j- 71 df 
will be, 

y — h + d (iA-|-Ai-}-A2 A3 ....iAw;. 



150 PRACTICAL MECHANICS. • 

The truth of this is readily established, for in tlie small 
interval between x =^ a and x = a -f d, it may be assumed 
that the function X remains constant. And because at the 
beginning of this interval X := A, and at the end X = Aj 
it may be assumed that this constant value is the arithmet- 
ical mean, or X = J A + i Aj. 

.-. For this interval X f ^ i= ^ (i A + J A,). For the 
next interval, comprised between 2; = ^ -f- ^ ^"d x^^ a -\- '^d^ 
it may in the same way be shown that 

X = iA, +iA, ; r. X[T = ^(i A, +iA,). - 

Adding this value of X [T to the preceding, it appears 
that the value of y from x = a, to a; = a + 2 d, is 

7/==5 + ^(iA + A, + I A,), 

and so on to the last. 

To apply this metliod to the determination of x and y in 
this case, it is known when p = tan. ^, then x = ; first 
take 

Now since the tangent p of the angle, which the curve 
makes with the direction of x, becomes gradually less and 
less, until, at the vertex of the curve it is nothing, and 
afterwards becomes negative, and goes on increasing in the 
descending branch ; make successfully p = tan, ^, tan, 
{0 — d), tan. {0—2 d), tan. (^ — 3 d), &c., until p = ; 
and then p = — d, — 2d. — 3d, &c., ad infinitum ; taking 
for d any number at pleasure : and the smaller it is taken, 
the more exact will be the discription of the curve. Cal- 
culate the values of A, Aj, A^, A,, &c., of the quantity 



P m — C, m 

when J) becomes successively tan. 0^ tan. {0 — d), tan.'{0 — 
2 rf), &c. Then putting successively n = 1, 2, 3, &c., ad 
infinitum, the values of x will be found that correspond to 
the angles where j) — tan. {fi — d), tan. {0 — 2 d), &c,, &c. 



PRACTICAL MECHANICS. 151 

Also, by the other equation 

_. f Pry 

y J Pm— 0,m ' 

the values may be found, which 

V 



P m — Cm 



acquires, when p becomes tan, ^, tan. {0 — d), tan. (0 — 2 d), 
cfec. ; proceeding in like manner, the values of y will be 
*<:nown, which correspond to the same angles, where p -^^ 
tan. (0 — d), tan. (0 — 2 d), &c. 

Hence the abscissas x, and the corresponding ordinates 
y of all the points of the curve, in which p becomes suc- 
cessively tan. {0 — d), tan. (0 — 2 d), &c 



NEW PRACTICAL SOLUTIONS OF IMPORTANT USEFUL PROB- 
LEMS, FROM COMBINING CONTINUED FRACTIONS AND 
DIOPHANTINE ANALYSIS. 

Question 1. It can be shown, that it is impossible to re- 
present the perpendicular on one of the sides from the 
opposite angle of an equilateral triangle in whole numbers, 
when the sides are expressed in whole numbers ; with this 
fact in view, let it be required to find a triangle, whose 
sides and a perpendicular may be expressed in whole num- 
bers, a.nd to have the three sides nearly equal. It is re- 
quired to have the difference between one of the sides, and 
either of the other two, less than any fractional part of 
either of the sides. 

Take any two numbers x and y, and according to Dio- 
phantine Analysis, 2xy will be the base, x"" — y"" the per- 
pendicular, and o:^ + y'' the hypothenuse of a rational right 
angled triangle. In an equilateral triangle, if the side = 
1, the perpendicular = \ ^Y. 

Let a;'' + 3/^ m^ 1, and 

2:^3/ :^i (3)^. 



152 PRACTICAL MECHANICS. 

.•. x + y^ \l + i {3)hU = 1.3660254 

X — y = I I — i (S)i U = 0.3660254 



.•. X = .8660254 
y zzi: .5000000. 



Now by continued fractions, the nearest whole numbers 
that will supply the required proportions may be found 



8660254 /, . . 
5000000 ^^-J"''^ 



5000000 ., , 
3660254 ^^- ''''''^ 



3660254 
2679492 



(2. third 



1339746 ... ,, 
980762 ^^- >^^^^ 



980762 ,^ .r.r 

717968 (^--^/^^ 



358984 .. . , 
262794 ^^' ^^^^^ 



St (2- --'^ 



96190 /, . ,^, 
70414 ^ • - ^ 



70414 xo • .7. 
51552 (^- ^^^^^^ 



25776 ,, , ,, 
18862 (^- ^'""^^ 



18862 .o 7 ,^ 
13828 ^ 6^6^"^^^>^ 



5034 {I > twelfth 



o^nr^ (2. thirteenth 



1880 ,, . , ,, 
1274- ^ * fourteenth 



1212 ^^' fift^'^th 



606 



PRACTICAL MECHANICS. 153 

The continued fraction will be 



^ + I 



+ 2 J_ i 

•^ 1 -4- * 



I 1 _4_ 1- 

» 2 J_ i 

•^ ^ 4- 4- I , 

+ ^ + i &c. 

summing five of these terms the fraction \\ will be found, 

then 

192 _|_ 112 _ 432 

19' — IP = 240 
2 X 19 X 11 ^ 418 

So that 241, 209, and 120, will form a rational right angled 
triangle, and two of these triangles placed together will 
form one nearly equilateral, the three sides will be respec- 
tively 240, 241, and 241, and the perpendicular will be 209. 
If eleven terms of the continued fraction be summed, f f f- 
is obtained, 

.•. 989' — 571' — 652080 

989 + 5n' =^ 1304162 

2 X 989 X 5n ^ 1129438. 

These numbers may be divided by two, and then the 
three sides of the required triangle will be 652080, 652081, 
652081 nearly equilateral, and the length of the perpen- 
dicular on 652080, will be exactly = 564719, a result easily 
verified. 

Ques, 2. It is not possible by plane geometry to divide 
an angle into three equal angles, yet, the tangent, secant, 
and radius of one third of any plane angle, may be found, 
in whole numbers, as near as we please, in the following 
way. Let the angle to be trisected 

^ 41°.. 44' .. 33". 72. 
3) 41°.. 44'.. 33". 72 
13 .. 54 .. 51 .24. 

The natural tangent of tliis angle is found from the tables 
to be .2477386. 



154 



PRACTICAL MECHANICS. 



Let 2;^ — 3/^ =-z: .247^386, and 

'^xy = 1.0000000 

.-. X = .7993650, y = 6254964. 

What follows is by continued fractions ; the alternate 
quotients are marked, ^rst secondj &c. 



1993650 first 
6254964 (1. 

1738686 third ' 
1038906 (1. 


6254964 second 
5216058 (3. 

1038906 fourth 
699180 (1. 


699180 fifth 
618252 (2. 

21528 seventh 
16206 (1. 


339126 

21528 . ,. 
sixth 

123846 '^^^• 
101640 


5322 


16206 eighth 
15966 (3. 



240 



377 
295 



+ * 



+ I 



+ 



+ »'^ + -f &c. 



377' 4- 295' 

377' — 295' 

2 X 295 X 377 



229154 

55104 

222430 



The three whole numbers 229154, 222430, 55104, forms 
a right angled triangle, one of whose angles will not differ 
from the third part of 41°.. 44'.. 33".72 by the third part 
of a second. 

Ques, 3. Find a body as near the form of a cube as we 
please, whose edges, diagonal of the base, and of two oppo- 
site corners, may be expressed in whole numbers ? 



Letx' + 7/' ^ (3)^ 
2 X y -- (2) ^ 



and 



PRACTICAL MECHANICS. 



155 



... ^ + 3,= |(3)i + {'2)i \i -Ut37n2 
x—y= -j (3) 4 — (2) i !■ i — .5637^05 



.♦. X — 1.1687108 
y = .6050003. 

To find the continued fraction ; 



1168^08,, . ^ 
6050003 ^^- ^"^"^ 



6050003 .J ^ 
5631105 ^^- '"^'''^ 



5631105 
412298 

1514125 
1236894 



412298 ., f .^ 
211831 (.^•/''^'^'^ 



(13. third- 



268931 (2- ^/'* 



134467 

8897 

45491 
44485 



(15. sixth 



ml (^- ---^'^ 



801 



1 + 



+ t'^ 



+ 1 



+ 



+ T^ 



^lllih eighth 
211 



+ &c., 



118.19 
6 118' 



^''■' &c. 
this continued fraction summed gives for eight terms 

11819'' + 6119'^ = 111118685, (A) 

11819'^ — 6119^ = 102258831, (B) 

2 X 11819 X 6119 -^ 144611284, (C) 

(B) is the length of each of the edges, (A) the exact length 
of the diagonal between two of the opposite corners. 
Further, if a four sided figure be constructed, each of its 
sides = (B), and (C), one of its diagonals, any one of the 



156 



PRACTICAL MECHANICS. 



four angles of this figure will not dilBFer from a right angle 
two seconds. All the lines are in whole numbers without 
decimals. If only five terms of the continued fraction be 
summed, the result = If , then 





85^ 


+ 


44^ 


= 


9161 




85^ 


— 


44^ 


zrr 


5289 


2 


X 44 


X 


85 


— 


1480 



There are four exactly equal square faces, the length of 
each side = 5289, the distance between two of the most 
remote corners o€ this approximate cube = 9161 ; the four 
sided ends are equal, the side of each end = 5289, and one 
of its diagonals = 7480. None of the angles of the ends 
differ i of a minute from a right angle. 

The Mc Galium Inflexible Arched Truss Bridge, ap- 
proaches nearer the standard of perfection than any other 
wooden bridge that have fallen under my notice, this fact 
is easily established by comparison and demonstration. 

Fig. 1 




Fig. 1, is what is known as the Burr Bridge, It is com- 
posed of lower and upper chords, and posts and braces. 
The posts are framed into the chords, and the braces are 
framed into the posts. Arches are placed on each side of 
the truss, securely fastened thereto, and extending below 
the lower chords, abut against the masonry. 

This form of truss was extensively used throughout the 
United States previous to the introduction of railroads. 
Many spans were of great length, and in cases where the 
arches were large, and the masonry sufiBciently permanent, 



PRACTICAL MECHANICS. 157 

this bridge was comparatively successful. Much difficulty 
was, however, experienced, by reason of the absence of 
counter braces. A moving load produced a vibratory and 
undulating motion, tending to loosen the connection of the 
timbers, which generally resulted in failure. 

Many of the first railroad bridges in this country were 
built upon this plan, but much greater difficulty was found 
in adapting it to the use of railroads, than had been pre- 
viously experienced in its use upon common roads. This 
difficulty arose from, 1st, the practical impossibility of per- 
fectly combining the action of the arch and the truss {Qd^oh 
system, of itself, being insufficient to carry the whole 
load) ; and 2d, the absence of counter braces. These de- 
fects, clearly apparent in their use on common roads, were 
greatly aggravated under the increased and concentrated 
nature of the weight, and the rapid transit of trains on 
railroads. It is true, they were obviated in part by adding 
largely to the amount of material in the structures ; but, 
as the difficulty was inherent in the plan, violent contor- 
tions in shape could not be prevented, and these in time 
caused failures. 

These remarks are intended to apply to spans of consid- 
erable length, as experience has proved that plans of even 
an inferior grade may be measurably successful in spans of 
ordinary length ; whereas, nothing short of the most judi- 
cious distribution of material will insure permanency, in 
cases where long spans are indispensable, and any arrange- 
ment which can be made permanent in the latter case, 
must certainly prove so in the former. 

It is worthy of remark here, that this particular combi- 
nation of the arch with the truss, is even now with some, a 
favorite idea, but it is believed that its warmest advocates 
will be geneially found among those whose opportunities 
for practical investigation have been limited, and that it is 
only necessary that the question be properly presented to 
them, to produce a change of views in respect to it. 

This partiality for the combination of the arch and the 
truss is attributable partly to the fact, that the simple truss 
nas in many instances failed, and as a last resort, the arch 
has been added, of such dimensions and strength, as to be 
competent to carry the truss and load, the truss serving 
only as a stiffener to the arch, while the latter, thrusting 



158 PRACTICAL MECHANICS. 

upon the masonry, has sustained the whole weight. Be- 
sides, to the casual observer, who has never studied bridge 
construction, this combination presents at least an appear- 
ance of great strength and solidity, which do not in fact 
exist. 

That the simple truss without the arch has failed in some 
instances, is unquestionably true ; but while many of these 
failures have been caused from inattention to, or ignorance 
of, the laws regulating the composition and resolution of 
forces, by far the greater nnmber have arisen from the in- 
ferior quality, or lack of the requisite amount of material, 
or from inferior workmanship. 

The acknowledged failure of the Burr Truss, as applied 
to railroad purposes, led to the invention of several other 
plans, all of which were based upon the abandonment of 
the arch, and were aimed at perfecting a truss, which, of 
itself, would be sufficient to meet the emergencies of the 
case. This was in pursuance of what was considered a 
very resonable hypothesis, namely, that one system properly 
proportioned, must prove much superior to any method or 
arrangement in which the attempt was made to combine 
two distinct principles, in their nature heterogeneous. 

Among the most prominent plans presented to remedy 
existing defects, was one invented by Col. Stephen H. 
Long. This plan of bridge was composed of lower and 
upper chords, posts and braces, similar in outline and gen- 
eral arrangement to the Burr Truss, but differing from it 
in detail. An efficient system of counter braces was intro- 
duced ; these were made adjustable by wooden wedges, as 
were also the sustaining braces, by means of which any 
desirable elevation or deflection might be given to the 
truss. This plan of truss was rigid to a degree not pre- 
viously attained : and to such an extent was this true, that 
when properly adjusted, no perceptible deflection was pro- 
duced by the passage of the load. 

It was, however, found difficult to keep it in adjustment, 
in consequence of the great shrinkage of the wedges and 
other timbers of the truss. 

The invention of what is known as the Howe Bridge, fol- 
lowed. In this, as in Col. Long^s bridge, the idea of com- 
bining the arch with the truss was originally abandoned, 
for reasons heretofore given, and it was believed that this 



PRACTICAL MECHANICS. 159 

simple form of truss would prove equal to any reasonable 
requirement. 

In the Howe Bridge^ the posts used in the Burr and Long 
bridges are dispensed with, and iron rods substituted, by- 
means of which any desirable camber may be given to the 
truss, thus overcoming the practical difficulty previously 
experienced in the adjustment of Col. Long's bridge, by 
the use of wooden wedges. 

This method of producing camber is certainly an im- 
provement upon the means adopted in the Long bridge, 
for that purpose, but is much inferior to the latter in its 
method of counter bracing, in that they are not adjustable, 
and perform a negative rather than a positive duty. 

The Howe Bridge is composed of lower and upper chords, 
braces and counter braces, vertical rods, and cast iron 
bearing blocks. The braces abut the bearing blocks, which 
pass through the chords in such a manner as to permit the 
rods to bear directly upon them. 

Spans of considerable length were built upon this plan, 
but experience proved that even this truss — like all others 
— had its limit, beyond which it could not be safely ex- 
tended. 

In the progress of railroad enterprises, in order to save 
large expenditures of money for masonry, longer spans 
than had been previously used became desirable, and in 
certain locations absolutely indispensable ; besides this, 
locomotives were largely increased in weight, to meet the 
demands of traffic, and furnish a more economical mode of 
working, and thus arose the necessity for the adoption of 
some other expedient to meet the increased requirements 
of bridges. As all had been done by way of improving 
this truss that mechanical skill could devise, and which an 
extensive practice had amply afforded, it became evident, 
that some radical change must be made in its arrangment, 
to enable it to meet the exigencies of the case. 

In this emergency the arch, heretofore condemned in the 
Burr TrusSf was again resorted to, for it had been proved 
from the experience which its use in that truss had afforded, 
that an arch of sufficient size abutting against permanent 
masonry, would place the truss in a position of secondary 
importance. 

It will be observed that the arch of the Burr Bridge, 



160 



PRACTICAL MECHANICS. 



b£> 1 




PRACTICAL MECHANICS. 161 

Fig. 1, abuts upon the masonry in precisely the same man- 
ner as the arch of what is denominated the Improved Howe 
Truss, Fig. 2, and the difference between the two consists 
simply in the mode of connection with the truss, and not in 
any change of principle, or method of action. 

It will be seen that the Burr arch is securely fastened to 
the posts and braces of the truss, forming a solid adjust- 
able mass. In Fig. 2, the arches are not fastened to the 
braces or rods, but have an independent connection with 
the lower chord of the truss, by means of rods radiating 
from the former to the latter. By this method it was sup 
posed that any desirable adjustment could be effected, and 
that the strain could be put upon either system, or equally 
upon each. 

This new arrangement, although plausible in theory, is 
found impossible in practice, for the following reasons : 

1st. The rods from the arch to the lower chord are of 
various lengths, consequently their contractions and expan- 
sions must vary proportionately. 

2d. Not a single rod in the arch is of the same length as 
those in the truss, hence the expansion and contraction of 
the rods in the truss will vary from that in each and all 
the rods connecting the arch with the lower chord. 

3d. This combination is exceedingly liable to maltreat- 
ment, from the careless or ignorant. 

4th. And even if it were everything in practice that is 
claimed for it in theory, (which is not the fact), it involves 
a constant expenditure for adjustment, which must continue 
during the existence of the bridge itself. 

The Burr Truss, Fig. 1, with all its defects, can be made 
superior by far to the Improved Howe Truss, Fig. 2. For in 
the former, there may sometimes be a yielding and com- 
pression between the parts of the truss and those of the 
arch, producing a certain degree of united action ; while 
in the Howe Truss, everything depends upon the length of 
the rods, which must always change with the temperature, 
and thus render an approach even to perfect adjustment, a 
matter of extreme delicacy. 

But in either Fig. 1 or Fig. 2, it is clearly evident that, 
in order to have a structure absolutely safe, the arch and 
the truss — each of itself, independently of the other — should 
be of sufficient strength to sustain the whole load, that tho 
strain may be borne alternately by each separate system. 



162 



PRACTICAL MECHANICS. 



In order to simplify and make clear the real points of 
difference existing in the combinations of the various plans 
of trusses, of the same general outline, it may be stated 
that the material composing any bridge trussj whether of 
wood or iron, or of both, is subjected either to tension or 
thru^t^ and it is upon the proper application of these ele- 
ments, together with a judicious distribution of the mate- 
rial, rather than upon any difference in detail, that the 
perfection of any bridge structure depends : this may be 
illustrated by reference to Figures 3, 4, 5. 




Figure 3 is the truss of the Burr Bridge ; in this the 
upper chord and braces are acted upon by thrust, and the 
lower chord and posts by tension. 

Figure 4 is the Howe Truss, without the counter braces ; 



PRACTICAL MECHANICS. 



163 



in this also, the upper chord and braces are subjected to 
thrust, and the lower chord and vertical rods are acted 
upon by tension. 

Fig. 4. 




Figure 5 is a plan of truss sometimes used, the counter 
rods being omitted ; in this the upper chord and vertical 
struts are subjected to thrusts, and the lower chords and 
diagonal rods are acted upon by tension. 

Fig. 5. 




Upon a comparison of these plans it will be discovered 
tliat the variations between the Burr Truss, Fig. 3, and the 
Howe Truss, Fig. 4, consists in the use of vertical rods and 
Dcaring blocks in the latter, instead of vertical posts in the 
former, both having precisely the same duty to perform. 



164 



PRACTICAL MECHANICS. 



It will also be seen that Fig. 5 varies from Fig. 4, in 
that the rods are placed diagonally instead of vertically, 
changing the element of thrust from the diagonal braces 
in the latter, to the vertical struts in the former, and trans- 
ferring the element of tension from the vertical to the diag- 
onal line. 

Much importance is sometimes attached to just such 
modifications in detail as exist in Figures 3, 4, 5, while the 
nature and intensity of the destroying forces are the same and 
equal in each. 

This has been proved by actual experiment, by the cele- 
brated engineer and bridge builder, D. 0. McCallum, as 
follows : Models were built, one on each plan, of equal 
length and height of trusses, containing the same sectional 
area and kind of material in chords and braces, and of 
equal perfection in details and workmanship, when it was 
found that the real difference in strength was unappreci- 
able, and it may be well to add, that any given amount 
placed upon each, in progress of the experiments, presented 
precisely the same characteristics and contortions in shape, 
until final failure took place. 

All bridges having their chords parallel, irrespective of 
the particular method adopted in combining them, and re- 




gardless of the amount of material used in their construc- 
tion, when loaded to nearly the point of fracture, present 
somewhat the sanae appearance, the greatest deflection being 



PRACTICAL MECHANICS. 



165 



invariably at points near the abutments. This will be un- 
derstood by the statement, that the vertical strain is in- 
creased, as the distance from the centre, to the ends of the 
trass ; at the centre the vertical strain is nothing, and at 
each end of the truss it is equal to one half the weight of 
the structure and its load. 

In point of strength, the arrangements Figures 3, 4, 5, 
are not superior to the simple combination Fig. 6. 







All bridges haying their chords parallel, exhibit tho 
same uniformity of action, and may be illustrated by refer- 
ence to Fig. 6 cr, in which A A, is the upper chord ; B B, 
tlie lower chord ; C, C, tension rods ; D, D, braces. 



166 



PRACTICAL MECHANICS. 



When a sufficient weight is applied co any truss of this 
outline, to cause deflection below the straight line, the up- 
per ends of the braces D, D, are made to approach each 
other, and the distance between the ends is diminished, and 
as the deflection increases, the upper ends of the braces D, 
D, will describe arcs a 6, of a circle downwards^ the radius 
of which being the length of the braces D, D. But when 








the upper chord is arched, as in Fig. 6 J, a sufficient weight 
will cause tlie braces D, D, to describe an arc upwards, re- 
presented by c d. Pig. 6 b. When tlic cliord c e becomes 



PRACTICAL MECHANICS. 



167 



straight, the arc will then be described downwards, as 
shown in Fig. & a. As an illustration of the McCallum 
Lijlexibk Arcfied Truss, see Fig. 7, in which A, A, is lower 
chord ; B, B, upper chord ; 0, C, tension rods ; D, D, 
braces ; E, E, struts, and W, weight. 




Upon an inspection of tliis figure, it will be seen that 
any deflection produced upon the centre of the arch, by 
means of the weight W, will cause the points B, B, to sep- 
arate, by thrusting outward, and in the direction of the ends 
of the truss, producing an upward movement of the upper 
chord, at the ends of the braces D, D, the latter describing 
arcs of a circle upward, and from thence will 1)0 couimuni- 



168 



PRACTICAL MECHANICS. 



cated by means of the tension rods C, C, to the centre of 
the lower chord, raising the latter at the point where the 
rods 0, C, meet. 

By removing the weight W. and inserting a Tcrtical 
strut at F, the upward movement of the chords will be 
arrested by the weight W. This peculiar action may be 
described as follows : 




Any deflection produced in the centre of the arch will 
cause an outward, and consequently, an upward force, at the 
upper ends of the braces, which, by means of the tension 
rods and strut, is transferred directly back to the under 
side of the arch, producing an upward force at the latter 
point, equal to the original downward force api^lied on top 
of the same. 



PRACTICAL MECHANICS. 



169 



This combination of forces is in agreement with a well- 
known law, namely, when two forces of equal powers of 
resistance are opposed to each other, a state of rest is pro- 
duced. 

For a further illustration of the action of this truss, see 
Fig. 8, in which A, A, are pieces of the lower chord, the 
centre being removed, B, B, upper chord, deflected by the 
weight W. C, C, are braces which pass through the lower 
chord, and rest upon the masonry. D, D, are tension rods. 
It will be seen that the ends of the pieces of lower chord at 




E, E, are raised considerably above a horizontal line. This 
upward tendency will continue until the upper chord be- 
tween B, B, is deflected below a straight line, when the action 
will be reversed. 



170 PRA^CTICAL MECHANICS. 

Figure 9 exhibits the forces at a state of rest, in which 
A., xl, are portions of the lower chord ; B, B, upper chord ; 
0, C, arch braces, which pass through the lower chord, and 
rest in the masonry. D, D, tension rods ; E, E, braces ; 
W, weight. 

It will be seen that the strain produced by the weight 
W, is transferred to the lower chord by means of thrust 
upon the braces E, E, to the points P, F, and by means of 
tension on the rods D, D, to the points B, B, and from 
thence it is brought upon the arch braces C, C, which rest 
upon the masonry. 

In this manner, a perfect equilibrium of forces is effected, 
as it is evident that the point G, cannot change position, 
unless the points B, B, are thrust outward towards the ends 
of the truss, which must raise these points, tliis being pre- 
vented by the strain upon the points F, F, communicated 
by the weight W, through the braces E, E. 

For a full plan of McCallum^s Inflexible Arched Truss, 
the reader is referred to Fig. 10. Upon inspection, it will 
be observed that the sustaining principle is very much in- 
creased toward the ends of the truss, not only by the 
addition to the amount of material at these points, but it 
will be seen also that the pannels become shorter as the 
vertical strain increases. The posts are placed upon lines 
radiating with the arch ; the braces form equal angles with 
the posts ; and in this way the latter are made to approach 
more nearly together toward the ends of the truss. 

The student has already had sufficient evidence of the 
great strength of this form of truss, and it has also been 
shown, that the tensile strain upon the lower chord is much 
less than in any other known plan. In fact, the latter may 
be entirely severed^ and the structure will still be competent 
to sustain a heavy load. In this, it differs from all other 
combinations. 

Upon referring to Figure 10, which represents a clear 
span of 180 feet, it will be seen that the arch braces which 
rest upon the abutments are extended to points on the arch 
about forty-seven feet from the abutments. From the top 
of each set of arch braces, running diagonally on each side 
of the truss, are placed heavy suspension rods, which are 
connected with the lower chords 12 feet further from th.c 
masonry. Thus the bridge seat is substantially transferred 




17] 



172 



PRACTICAL MECHANICS. 







to a point 47 feet towards the centre of the bridge, reduc- 
ing- a span of 180 to 86 feet, so far as the tensile strain upon 
tlie lower chord is concerned. 

For this intermediate space of 86 feet, the arch beam is 
of sufficient strength to sustain tlie whole load, if required. 



PRACTICAL MECHANICS. 173 

Strength, however is not all that is required, for a Rail- 
road bi'idge especially, subject as it is to a moving load, 
there must also be rigidity, stiffness, freedom from vibration. 
A bridge may be strong yet flexible, rigid yet weak ; in 
fact, flexibility is incompatible with durability, the struc- 
ture should be prepared at all times to receive its load, and 
should not be permitted to change shape in the slighest 
degree, by its passage over it. 

To produce this result, an effective system of counter 
braces, is indispensible. 

The proper office of counter braces is frequently misun- 
derstood, as is evident from the manner of their application 
in many cases in which they are used as check braces only, 
having a negative rather than a positive action ; this may 
be readily shown. When the load is applied, the truss is 
deflected in consequence o\ the yielding of the braces ; 
this has the effect of shortening the diagonals in the direc- 
tion of their length, while the diagonals in the direction 
of the counter braces are correspondingly lengthened ; this, 
will leave a space between the ends of the latter, and the 
bearing block in the lower chord. 

When the truss is in this condition, if wedges are insert- 
ed between the ends of the counter braces and the lower 
chord, in such a manner as to fill up the whole space, it is 
evident that the weight may be removed without at all 
affecting the shape of the truss, the deflection originally 
produced by the weight, being maintained by the counter 
braces, the strain upon the sustaining braces and other 
portions of the truss remaining precisely the same as when 
the weight was suspended. 

Now suppose tlie original weight to have been 200 tons, 
it is evident that as soon as it is removed, each counter 
brace will be subjected to an upward thrust, easily found 
from its position ; the sum of all the thrusts making 200 
tons. 

Now let there be a smaller load applied, tliis load will 
not produce any additional strain upon any portion of the 
truss, nor will the deflection be increased in the slightest 
degree ; the only effect produced by suspending the latter 
weight, will be the relief of the counter braces, equal to 
the diSerence between the first and second weights. 

The inventor has found it very difficult to explain this 



174 PRACTICAL MECHANICS. 

clearly in the course of conversation with some individuals, 
from the fact that weight and strain were confounded. Now 
it is true, when the original weight was applied of 200 
tons, the abutments were loaded with just 200 tons more 
. than previously, and the truss was also loaded with ten 
tons more ; but when the wedges were driven, and the 
weight removed, wliile the abutments were relieved of 200 
tons pressure, the truss still retained the original strain 
produced, the weight being required to produce the strain, 
the latter remaining after the former has been removed. 

In order to make a practical application of the above, 
the following method of adjusting the Inflexible Arched 
Truss, is submitted. When these bridges are raised, it is 
usual to load them with a train of locomotive engines, at- 
tached closely to each other, and that greater weight may 
be obtained, the tenders are sometimes detached, and the 
bridge covered with engines only ; with this load, the lat- 
ter is strained down to a perfect bearing in all its parts ; 
by this means the whole structure is more or less deflected, 
while the counter braces are hanging loosely in their places ; 
if, therefore, when the bridge is in this condition with its 
load, the counter braces could be lengthened with consid- 
erable force, it would not recover its original shape upon 
removal of the load, but would be held down by the action 
of the counter braces to very nearly the same position as 
when loaded. In this plan of bridge, the lower ends of the 
counter braces rest in iron stirrups, which are attached to 
the vertical ties or posts at a point near the lower chord 
by means of castings and nuts, by which they may be 
lengthened several inches, in, this manner they are made to 
perform a positive duty. When the bridge is adjusted as 
above, it is clear that a less load than that originally ap- 
plied cannot produce any deflection whatever, the only 
effect of the passage of a train over it will be to relieve 
the counter braces, and will not add a pound pressure upon 
any timber of the trusses. 

In the arrangement of any bridge truss, the attainment 
of the following requisites is desira,ble : 

Fii^st, Such equilibrium of forces as will produce uni- 
formity of action. 

Second, Such quantity and distribution of material, as 
will insure a large surplus of sustaining principle thereby 
guarding the structure against accident. 



PRACTICAL MECHANICS. 175 

Thirds Perfect rigidity, that the combination in all its 
parts may have permanency equal to the durability of the 
material composing the same. 

Fourth, The arrangement of the parts should be such as 
to be free if possible, from the iiecessity of adjustment. 

The McCallum Inflexible Arched Truss meets all these re- 
quirements. 

PROBLEM. 

Let it be required to find the equal weights w, w, Fig. 
11. Kept in a state of rest by a single weight W, which 
has caused the arc q C a to assume the chord q Db, the 
rigidity of the arc being neglected. 

Let R = the radius of the arc q G a, and put 2 c = its 
length ; then the chord q a and versine D are readily 
calculated. 

Let g a = 2 ri and C D = 2 e. 

Let f-=^Aq = Ap = Ba=^'Bb and 

g zzzi B n =^ B w = A 5 i^ A ?*. 

Also, let A B = 2 A 

2 k — 2d , ^ 

> ^ ^h-d, 

h — d 



COS. 6 J putting 6 for the angle a B n = ^ A 5 
h — c. 



/ 

2 A — 2 c 



2h — 2c= AB—qb. 
Putting (f for the angle bB n ^:= q As, 
h — c 

~ zzi: cos, (p. 

.•. <p — 6 = angle aB b =i nBm i:= s Ar, 

.•, mn •=^ g sin. (cp — 6). 



176 



PRACTICAL MECHANICS. 




PRACTICAL MECHANICS. 



177 



Then according to the principle of work, 

2 g sin. (qp — 6) IV -z^ 2 6 W. 



w :r^ 



^Y 



g si7i. (go — 6)' 



IRON SUSPENSION AND TRUSSED BRIDGE AT HARPER's 
FERRY — DESIGNED BY WENDEL BOLLMAN. 

The span of the Iron Suspension and Trussed Bridge, at 
Harper^s Ferry, ib 124 feet beiTveen the abutments. The 
length of cast iron in the stretcher is 128 feet. The weight 
of cast iron is 65137 lbs. ; weight of wrought iron 33527 
lbs., making the total weight of cast and wrought iron 
98664 lbs. 

COHESIVE STRENGTH PER SQUARE INCH OF CROSS-SECTION. 



KIND OF MATERIALS. 



Ca5t Steel, 

Blistered Steel 

Steel, Sheer, 

Iron, Swedish bar, 

'' Kussian, 

" English, 

" common, over 2 inches square 

•• sheet, parallel rolling, 

'• at right angles to roll, 

Cast iron, good quality, 

Platinum, wire 

Brass, common, 

Wood. 

Ash. 

Beach, 

jBox, 

iCedar, 

'Mahogany, 

I *' Spanish, 

I Oak, American white, 

" English; 

Pine, Norway, 

Walnut, 

Hemp ropes, good, 

Iron chain, 

•' with cross pieces, 



JUST TEAR ASUXDER. 



Pounds. 



134256 

133152 

128632 

65000 

59470 

56000 

36000 

40000 

34400 

45000 

53000 

4500 

16000 
11500 
20000 
11400 
21000 
12000 
11500 
10000 
13000 
7800 
6400 
65000 
90000 



Tons. 



59.93 

59.43 

56.97 

29.2 

26.7 

25.0 

16.00 

17.85 

15.35 

20.05 

23.6 

20.05 

7.14 
5.13 

8.93 

5.09 

9.38 

5.36 

5.13 

4.46 

5.8 

3.48 

2.86 

29 

40 



\7ITH SAFETY. 



33600 
33300 
32160 
16260 
14900 
14000 

9000 
10000 

8600 
11250 
13250 

1125 

40©0 
2875 
5000 
2850 
5250 
3000 
2875 
2500 
3250 
1950 
2130 
21600 
30000 



Tons. 



14.98 
14.86 
14.24 
7.3 
6.7 
6.25 
4.0 
4.46 
3.84 
5.00 
5.9 
5.0 

1.87 
1.28 
2.23 
1.27 
2.34 
1.34 
1.28 
1.11 
1.45 
0.87 
0.95 
9.38 
13.4 



178 



PRACTICAL MECHANICS. 



Fig. 12, is an elevation of part of the side, showing one 
pier, part of the cast iron stretcher. The cap is removed 
from the pier to show how the rots are secured. 

Fig. 12. 




Fig. 13, is an elevation of a pier and four pannels out 
of eight of which the bridge is composed. The s.vstem of 
arranging the braces and connecting rods is exhibited in 
this fis-ure. 



PRACTICAL MECHANICS. 



179 




180 



PRACTICAL MECHANICS. 



Fig. 14, is cross section, showing the floor bracing, and 
the position of the rails. Fig. 14, also shows in section, 
the roof and posts. 

Fig. 14. 




Fig. 15, sliows a plan of tlic flooring of the bridge, the 
positions of tlie rails and floor bracing. 

Fig. 16. shows two posts part of the stretcher, and the 
diagonal rods in one of the pannels. 

Tbe wrought iron requires little workmanship ; the rods 
from the centre to abutments having but an eye at one and 
a screw at the other, end ; witli a weld or two between, 
according to length. The long counter rods liave two 



PRACTICAL MECHANICS. 
Fig. 15. 



181 




182 PRACTICAL MECHANICS. 

knuckles and one swivel for adjustment of strain, and con- 
A^enience in welding, as well as in raising the whole. 

The cast iron stretcher is octagonal without, circular 
within, and averages one inch of metal. It is cast in lengths 
according to the length of panel, and jointed in the simplest 
manner ; — at one end of each length is a tenon, at the other 
a socket. The latter is bored out, and the tenon and its 
shoulder turned off in a lathe to fit the socket ; thus, when 
thoroughly joined, to form one continuous pipe between 
abutments. The ends of the sections of cylinders, in- 
serted to those contiguous, are slightly rounded, to allow 
a small angular movement without risk of joint frac- 
ture. 

A cast iron plate or washer, sets on a bracket cast with 
each abutment end of stretcher, and at right angles to the 
centre acting rods. The tension bars are passed through 
this washer to receive a screw nut for the erection and ad- 
justment of the system. 

Tlie stretcher or straining beam, the vertical posts, and 
suspension bars compose the essential features of the bridge ; 
each post being hung by two bars from both ends of the 
stretcher independently of all the others ; and each post 
and pair of tension bars forming with the stretcher a sepa- 
rate truss. 

This system, perfect in itself, is additionally connected 
by diagonal rods in each panel ; also by light hollow cast- 
ings, acting as struts. The diagonal side rods might be 
sa4'ely dispensed with, for the peculiar merit of the truss 
is its perfect independence of such provision. They are 
therefore used as a safeguard only in case of the fracture 
of any of the principal suspension rods. 

By this combination of cast and wrought iron, the former 
is in a state of compression, the latter in that of tension ; 
the proper condition of the two metals. It unites the 
principles of the Suspension and of the Truss Bridges. 
Each bar performs its own part in supporting the load in 
proportion to its distance from the abutment"; so that the 
entire series of suspending rods transmits the same tension 
to the points of support as would be equally transmitted 
from thence to the centre of the bridge. 

This bridge, it will be seen, is composed of seven inde- 
pendent trusses, which transfer the weight concentrated on 



PRACTICAL MECHANICS. 183 

each floor-beam directly to the abutments, without aid from 
any other connection ; and not from panel to panel as in 
general use. 

The strain on cast and wrought iron is wholly in direct 
line ; and the result, the least quantity of metal is required 
to carry a given weight. The weight of bridge and load 
has a vertical pressure on the piers, towers, &c., the only 
horizontal thrust being from the expansion of iron, which 
is accommodated by rollers, sliding of abutment bracket 
over its pedestal, or by other means : the necessary dimen- 
sions of masonry may therefore be most moderate". 

It is evident, from an inspection of the drawing, that no 
chord is requisite at the bottom of the truss to resist ten- 
sion ; the only advantage of that employed is to regulate 
the movement produced by expansion, in the performance 
of which agency the resistance is one to compression. 

Although the abutment bracket casting and its pedestal 
were so constructed as to admit of accommodation to ex- 
pansion, by rollers, yet such contrivance was omitted with 
the view of fully testing the effect of greatest expansion 
throughout the system. 

It is now ten months since this bridge was erected at 
Harper's Ferry ; during which time it has been exposed to 
extremes of cold and heat, and to an average run of twenty 
trains daily. 

From the closest inspection, we find that the extreme 
expansion measures, as near as possible, five-sixteenths of 
an inch on each tower, or five-eighths in the entire length, 
128 feet of stretcher ; and without the slightest percep- 
tible derangement of masonry ; the dimensions of which 
lire 4 feet square of base, 12 feet high, and 2 feet 9 inches 
'at top. 

While on the subject of expansion, it may be well to 
notice the effect from difference in expansion of the rods. 
xVt the first point of suspension, or where the longest and 
shortest rods meet, the counter rod is about four and a half 
times longer than the acting rod ; and the expansion of 
the counter is four and a half times that of the acting rod. 
But there is also a proportionate difference in the lengths 
of stretcher from the point directly over the centre of con- 
nection to the extremities of these rods. This has been 
practically proved in this bridge. 



184 PRACTICAL MECHANICS. 

Tlie suspender bolt, when the expansion is extreme or 
five-eighths of an inch in length of stretcher, exhibits a 
motive difference of three-sixteenths toward the short 
or acting rod ; which difference is provided for, as seen by 
slot-dotted in Elevation, where the vertical suspender bolt 
moves to accommodate any such difference, and to give 
that proportion of weight to each rod according to tlie 
angle. 

It affords easy access for repairs ; for instance, should a 
new floor beam be required, it is but needed to slacken the 
horizontal rod and the keys in longitudinal strut, remove 
the washer under point of suspension, and let down the 
beam to be replaced ; which can be done without trestling 
up any part of the bridge. 

In case of fire, the floor may be entirely consumed with- 
out any injury to the side truss. 

The permanent principle in bridge building, sustained 
throughout this mode of structure, and in which there is 
such gain in competition with every other, viz : the direct 
transfer of weight to the abutments, renders the calcula- 
tion simple, the expense certain, and facilitates the erection 
of secure, economical, and durable structures. 

Trial made on the 1st day of June 1852, to prove the 
capability of this Bridge, 

Three first-class tonnage engines, vrith three tenders, 
were first carefully weighed, and then run upon the bridge, 
at the same time nearly covering its whole length, and 
weighing in the aggregate 273,550 lbs., or 136|^ff^ tons 
nett, being over a ton for each foot in length of the brid^'e. 

This burden was tried at about eight miles per hour, and 
the deflections, according to guages properly set and reli- 
able in their action, were at centre post, If ^', and at the 
first post from abutment nine-sixteenths of an inch. 

From this test it is found, that the load did not cover 
the entire length of bridge by about 13 feet, yet the excess 
of weight iu the middle, and at a speed of about eight 
miles per hour, produced no greater deflection than If of 
an inch at the centre post, and nine-sixteenths of an inch 
at tlie first point from abutment. 



PRACTICAL MECHANICS. 185 



EXAMPLES FOR PRACTICE. 



PROBLEM I. 



To find the difference between the logarithms of two 
consecutive numbers without knowing the logarithms of 
the numbers themselves, and to calculate the logarithm of 
any given number. 

RULE. 

When the numbers are greater than 100, divide .868589 by 
double the lesser number plus one^ and the quotient will be 
the difference of the logarithms of the given numbers, 

EXAMPLES. 

Question 1. What is the difference between the logar- 
tms o" " " 
either 1 



ithms of 125 and 126, without finding the logarithms of 



2 X 125 + 1 ::^ 521. 

.868589 



251 



^ .003461 



So that the difference of tlie logarithms of 125 and 126 
is = .003460. 

Ques, 2. What is the difference between tlie logarithms 
of 126 and 127 ? 

.868589 

= .003433 



253 

Ques, 3. What is the difference between the logarithms 
of 127 and 128 ? 

.868589 

— .003406. 



255 



186 PRACTICAL MECHANICS. 

Hence, the difference between the logarithms of 125 and 
128 becomes known, without knowing the logarithms of 
either 125 or 128, or of the intermediate numbers. 

.003406 
.003433 
.003461 



.010300 



The difference between the logarithms of the numbers 
125 and 128, is found to be .010300, which I will put = n 
in the next example. 

Ques. 4. Required the logarithms of the numbers 2 
and 5? 

5^^ri?-V:^125 



(^) 



2' = -^ 128 

log. 128 — log. 125 = 71= .010300 

.-. log. (2)^— log. (-^) = n 

.•. 7 log. 2 — 3 log. -^ =: n 

.•. 1 log. 2 — 3 log. 10+3 log. 2 --= n. 

10 log. 2 = 3 log. 10 + ^i == 3.010300 

.-. log. 2 = .3010300. 

This method for calculating the logarithm of 2 is origi- 
nal. Hence, the log. of 5 

log. 10 = 1.000000 
locr, 2 '= .301030 



log. 5 = .698970 



PRACTICAL MECHANICS. 187 

7 log. 2 = Log. 128 = 2.107210 
3 log. 5 = log. 125 — 2.096910. 

log, 125 = 2.096910 

.003461 difif. found in example 1 



log. 126 rr: 2.100371 

.008433 difif. found in example 2. 



log. 127 = 2.103^04 

.003406 difif. found in example 3. 



log. 128 = 2.107210 



The logarithms of 2, 200, 2000, &c., and of 5, 500, 5000/ 
&c., also become known, 

Ques, 5. Reouired the logarithms of 101, 2001, and 
5001 ? 

.868589 ^,,o^. 

-To— - ■'''''' 

In this way the difference between the logarithms of 
100 and of 101 is found, 

.•. log. 101 = 2.004321. 

^'S^ = — ■■ 

but the logarithm of 2000 = 3.301030. 
Therefore, log. 2001 = 3.301247. 

4^-^ ^ .000087 
10001 

log. 5000 = 3.698970 



3.699057 ^-- log. 5001. 



188 PRACTfCAL MECHANICS. 

ques. 6. The logarithm of 125 has been found = 2.096910, 
find the logarithms of 123, 124, 41, and 31 ? 

2 X 123 + 1 1:1= 247. 

.868589 



24 1 

.868589 
■~249 



:^ .003517 
— .003488 



Differences 
required. 



log. 125 — 2.096910 

.003488 difference. 



log, 124 = 2.093422 

.003517 difference. 



log. 123 ^ 2.089905 



41 X 3 = 123 
31 X 4 = 124 

Log. 124 z= 2.093422 
log. 4 =: .602060 



Log. 31 -^-- 1.491362 



2^°== 1024 

Log. 1024 = 3.010300 

.868589 



2049 



.000424 



Log, 1025 = 3.010724 



Log, 1025 — log. (5 X 5 X 41) == 2 log. 5 + log. 41 
.-. log. 41 ^-1= log. 1025 — 2 log. 5. 



PRACTICAL MECHANICS 189 

But the logarithms of 5 and 1025 are already found, 
consequently 

Log. 1025 = 3.010724 
2 log. 5 — 1.397940 



log. 41 =: 1.612784 



Log. 123 ^ 2.089905 
log. 41 = 2.612784 



log, 3 = .477121 



Log. 126 =:=: log. (9 X 2 X t) =z 2.100371 

log. 9 = .954242 



1.146129 
log. 2 = .301030 



log, 7 = .845098 



In the same way may the logarithm of any given num- 
ber be found. 



MECHANICAL EXAMPLES. 

RULE T. 

If a force be exerted through a given space, and the power ex- 
pended ascertaiiied, then by quadrupling the force and doub- 
ling the space, it vnll be found that an expenditure of eight 
times ^he power will i^esuli J this is true. 



190 PRACTICAL MECHANICS. 

RULE II. 

The resistance of a cylinder which moves in a fluid in the 
direction of its axis, is equal to the weight of a column of 
fluids whose base is equal to that of the cylinder^ and its alti- 
tude equal to the height through which a body must fall in 
vacuo by the force of gravity^ to acquire the velocity of tlie 
moving body ; this rule is also true, 

I will iSrst find what resistance would be encountered 
by a plane of a foot square, urged through water at the 
rate of three miles an hour. 

5280 X 3 
—^ — = 4.4 feet = space passed through in a second. 

In this case the well-known formula 

v' (4.4)' 3 

_ — gives — \ — ^— - — 3 ft. or — — of a foot, the height from 
2g 2 X o2i 10 

which a body must fall to acquire a velocity of 4.4 feet a second. 

62.5 X .3 == 18.75 lbs 

This result may be tested by rigorous experiments made 
in France by D^Alembert, and other philosophers. They 
found that a surface of 1.1363 American feet with a velo- 
city of 2.7263 feet per second, had a resistance equal to 
8.234 lbs. avoirdupois, which is equal to 7i lbs. to the 
square foot. 

Then if it be assumed, for it is near the truth, that re- 
sistances vary as the squares of the velocities, 

lbs. lbs. 

(2.7263y : 7i : : (4.4)^ : 18.8 
almost exactly what is found by theory and Rule II. 

Let us sec how Beaufoy's experiments agree with theory : 
— the experiment was made on a foot square, the velocity 
was 9.2 miles an hour, and the resistance 203.79 lbs. 



PRACTICAL MECHANICS. 191 

lbs. lbs. 

.•. (9.2)' : 203.79 : : (3)'^ : 21.0 
Theory makes the result 18.8 lbs. 

If, for the sake of argument, we take 20 lbs. for the re- 
sistance, at a velocity of three miles an hour 

lbs. lbs. 

3' : 20 : : 6' : 80 

which shows that 80 lbs. will be the resistance to a square 
foot at six miles an hour. The question in dispute is this ; 
if the gravity of a weight of 20 lbs. is the measure of the 
resistance to a plane a foot square at the velocity of three 
miles an hour, and the gravity. of a weight of 80 lbs. is 
the measure of resistance to the same plane at a velocity 
of six miles an hour, is the power necessarily expended 
per hour in the latter case eight-fold or four-fold, what it is 
in the former ? 

A quadrupled resistance is already predicated, which 
amounts purely from doubled velocity. Now, let the plane 
of a foot square be expanded to an area of 4 square feet, 
its resistance would then also be increased four-fold ; but 
that would arise from increased surface. The error arises 
from considering these resistances as identical initial resist- 
ances, and then putting the term pressure as the represent- 
ative of both. This is a fallacy. The one is absolute, the 
other virtual. Are they equal in effect ? with regard to 
time they are. With regard to space they are not ; but 
the cube theory requires that they should be with regard 
to space. To make this clear, imagine a ptilley, the 20 lb. 
weight suspended over it by a cord attached to the one foot 
plane, a well being under the weight ; also, imagine that 
the plane is in the water, at a convenient distance, that the 
cord weighs nothing, and that there is no rigidity or fric- 
tion. Then the gravity of the weight pulls the plane 
through the water, and its velocity becomes constant when 
it moves at the rate of three miles an hour. Detach the 
20 lb. weight and the one foot plane, and substitute the 80 
lb. weight, and the plane of four feet.' The velocity of 
this plane also becomes constant at three miles an hour, 
being an actual four-fold resistance, and a four-fold power 



192 PRACTICAL MECHANICS. 

to sustain it. Take off the plane of four feet, attach the 
plane of one foot, and see what the 80 lb. weight will do 
Avith that. Its velocity becomes constant and uniforni at 
the rate of six miles an hour. Time, that all important 
clement, must now be introduced. The 20 lb. weight 
draws the plane of one foot three miles an hour. The 80 
lb. weight draws the plane of four feet the same distance 
in the same time. Here, then, is an actual four-fold resist- 
ance, and a four-fold expenditure of power. But the plane 
of one foot can have but one-fourth the initial resistance 
of a plane of four feet. Therefore, to expend as much 
power upon the one foot plane as upon the plane of four 
feet, it must be dr^wn through the fluid double the dis- 
tance. That is, it must be drawn through six miles instead 
of three in the same time : and tlie expenditure of power 
being four-fold for a doubled velocity, it is as the square 
of the velocitv and not as the cube. 



FOR PADDLE-WHEEL STEAMERS. 

Let a ::i^ area of all the floats immersed, in square feet. 

h = the immersed angle of the paddle-wheels, at the centre 
of pressure. 

c -^^ radius from the centre of motion to the centre of pres- 
sure of the floats. 

d = acting area in square feet. 

d zziz a COS. — — 

Suppose the area of all the immersed paddles a = 294.6 
square feet, the angle b = 47° ; what is the acting area ? 

^^ = 15° .. 40' : Natural cos. r— 9628490 



3 

For practical purposes .963 will be sufficiently accurate. 

294.6 X .963 = 283.6998 square feet acting area. 



PRACTICAL MECHAA:l>>. 193 

Put e ^=z the length of the loadline of the vessel, in feet. 

/ =z:z the great immersed section, in square feet. 

g z=z area of resistance in square feet. 

h =: displacement of the vessel in tons. 

i =z horse power required to propel the vessel j miles an 
hour, (statute miles.) 

k = co-efficient of the vessel. 

I == resistance of the vessel in pounds. 

m ==: tabular index corresponding to k. 

n = slip of either propeller or paddle-wheels. 

= the pitch of the propeller. 

f = the number of revolutions a minute. 

^ EXAMPLES. 

Ques. 1. Required the area of resistance of a vessel 
when e = 260 feet, = the length of the loadline ; the great- 
est immersed section 500 square ieei = f ; the displace- 
ment of the vessel 2500 tons = h, 

» 

35/^ _ 35 X 2500 _ 
^, = "77" - 500 X 260 ^ '^ 

Opposite .67 in the following table, k, will be found = 
1.76, the co-eflficient of the vessel. 

f ^ •'500') ^ 

Then^ = V j^j^ == s/ 500 XI .76 + (260 7 ^ ''^ 

square feet, the area of resistance required. 



19J: 



PRACTICAL MECHANICS. 



TABLE. 



i 
in 


k 


711 


k 


.50 


1.31 


.71 


1.52 


.51 


1.37 


.72 


1.43 


.52 


1.45 


.73 


1.36 


.53 


1.51 


.74 


1.29 


.54 


1.56 


.75 


1.21 


.55 


1 62 


.76 


1.13 


.56 


1.66 


.77 


1.05 


.57 


1.71 


.78 


.978 


.58 


1.76 


.79 


.903 


.59 


1.83 


.80 


.835 


.60 


1.89 


.81 


.763 


.61 


1.93 


82 


.691 


.62 


1.98 


83 


.625 


.63 


2.01 ! 


84 


.559 


.64 


1.97 


85 


.496 


.65 


1.91 


86 


.433 


.66 


1.84 


87 


.379 


.67 


1.76 


88 


.326 


.68 


1.72 


89 


.277 


.69 


1.64 


90 


.229 


.70 


1.59 


95 


.025 ! 



Qxies, 2. Required the resistailce of the vessel in pounds 
^— /, when she is running at the rate of 10 miles an hour 
--=^ j ; the area of resistance := 40 feet = g, found as in 
the last example. 

/ ^ 4.22 X gj' = 4.22 X 40 X 100 — 16880 lb- , 
tlif resistance of the vessel in pounds. 

The resistance in pounds may be found by the following 
formula nearer the truth 

Log. I — .B38468 -f- iog. g -j- 1.15 log. j. 



Log. J 



log. 10 --= 1.000000, 



PRACTICAL MECHANICS. 195 

1.75 log, j — 1.750000 

log.g = 1.602060 

Constant = .633468 



Log. 9672.3 '-^ 3.985528 



The first rule gives the resistance =: 16880 lbs 

Ques. 5. Required the horse power i, when the speed is 
e(][ual 11 knots an hour = j ; and the area of resistance 
= 33 square feet = g, 

. . j' X g 11^ X 33 ^^^ . 

^ =-^ — gg — — ^ = 49^ horse power. 

The horse power may be found with greater accuracy 
by the succeeding logarithmic expression 

Log. i == 2.75 log. J -f log. g — 1.944483 

' Log. J ^ 1.0413937 ; log.g = 1.518514 

2.75 X 1.041393 = 2.863830 

log.g = 1.518514 



4.382344 
Constant — 1.944483 



Log. 274.07 -^ 2.437861 



The horse power according to this latter method = 
274.07 ; there is a discrepancy in these results also ; the 
cube of the speed is employed in one case, and 2f power 
ill the other 

- — — — ^ y put into a logarithmic form 

oo 

gives the latter rule. Experiment has not yet settled this 
point in steam navigation, simple as it is, some employ the 



196 PRACTICAL MECHANICS. 

square of the velocity, others the cube, and a third party 
the 2f power. Government experiments, both in this coiub 
trv and in England, are generally government jobs. Ln- 
o'ineers are selected to superintend them upon the principal 
that determined who should be the village schoolmaster ; 
one of those worthies, was chosen to fill the important 
office of village teacher, merely because he had a large 
family and a wooden leg. 

PROBLEM 4. 

To find the slip of Propeller or Paddle-wheels, when the 
acting area, and area of resistance are given. 

If the area of resistance = 49 feet, = g ; and the act- 
ing area = 225 feet = d ; required the slip = n. 

773- _ 343 ^ ^43_ 

^ ^ -J^^f-Zf-jrjT ~ 343 + 33T[5 ^3718 * 

PROBLEM 5. 

To calculate the power and find the properties of Pad- 
dle-wheels and Screw Propellers. 

The pitch of a propeller = 33 feet ; and makes 42 revo- 
lutions a minute, the slip = .35. What is the speed of the 
vessel ? 

.^^ ^ ^^ (1 — .35) -^ \\A knots. 

88 ^ ^ 

Put =-. the pitch ; p = the revolutions a minute ; and 
n = the slip ; then generally 

^-— g— (1--); 

— 88^ __ 

•'• ^ ■" "^X (1-^) ' 
88 j 



n = ^ — 



p 



PRACTICAL MECHANICS. t97 

Let the speed be 11.7 knots the hour, = j ; the slip = 
.35 = n; the pitch 33 feet = o; required the number 
of revolutions = p. 

88 i ^ 88 X 11.7 _ 

^ X {l—n) 33 X (1 — .35 "~ 

the nuniDer of revolutions a minute. 

Required the slip, when the speed is 11.7 knots per hour, 
revolutions 48 per minute, and the pitch = 33 feet. 

. =. 1 - -^^- = 1 ^ '^ X ^^-^ - 35 
X Jp 33 X 48 

PROBLEM 6. 

Any four of the five following quantities being given to 
find the fifth, namely, the radius of a paddle-wheel from 
tlie centre of motion to the centre of pressure of the floats, 
= c ; the .slip, = n ; the immersed angle of the paddle- 
wheels at the centre of pressure = b ; the statute miles 
per hour = j ; and the revolutions a minute = p. 

From the general expression 

c p COS. i b , 

any one of the quantities is easily deduced, for 

i>=-— ^i^ ^.n ^i- 

(1 — n) c COS. I ' pc COS. i b 

Suppose the immersed angle of the paddle-wheels at the 
centre of pressure = 54°.. 33' = 6. The slip = .38 = ^ ; 
the revolutions per minute = 1& == p ; and the radius of 
the paddle-wheel from the centre of motion to the centre 
of pressure of the floats = 21 feet ; required the speed 
= i ? 

One third of 54"^ .. 33' = 18° .. 11', the natural cosine of 
winch IS = .9500629, for practical purposes .95 will be 
sufficiently near tlie truth. 



198 PRACTICAL MECHANICS. 

^ _cpcos_^J_ 21 X 16_X^ _ 3 ^ 

14.136 miles an hour. 

Again, let the immersed angle = 54° *. 33' ; the revolu- 
tion per minute = 16 ; the speed 14.136 miles an hour ; 
the radius of the paddle-wheel from the centre of motion 
to the centre of pressure = 21 feet ; required the slip ? 

16 X 21 X .95 
the required slip. 



CALCULATIONS RESPECTING THE ANTI-SLIP 
SCREW PROPELLER. 

Put ^ ^=: :^ the pitch at the periphery = D E. 
r = :J the pitch at the hub = R S. 
s := the assumed slip in a fraction = H R, of 5^. 

••• q -^ rJ^ '^• 

Put t :^ extreme radius =: Q D. 

u = diameter of the screw = 2t = FI. 

V =^ pitch of the propeller at the periphery = 4 ^ . 

w = angle of the blades at the periphery 1::= K Q T. 

w , i-= angle of the blades at the hub. 

J M = X zz^L ^ pitch of the propeller at the centre of effort of 
the blades. 

B J ^^ y = .725 t, the radius at the centre of effort G, —- 

OGt ::^ QM. 

. •. The actual pitch at the centre of effort ::^- 4 x. 

z 1=^ horse power required to drive the propeller and 
oc revolutions a minute. 



PRACTICAL MECHANICS. 



199 




•^ 
C 






200 PRACTICAL MECHANICS. 

B 1^ number of blades. 

y ^zr length of blade parallel with the centre line — U V 

d z:=: the breadth of the propeller blades over the edge 
between the corners X z 

s z=z the circular arc in the angle ^ z^r W Y. 

6 znz the projected angle of the blades 

A =:= the true inclined surface of the blades 

A = the projected area of the blades. 

A 2 ::::= tne actlug area of tne propeller. 

Let 7T = the circumference of a circle diameter ^=: 1 . 

Cot. w = :=r — ^ — = the cotangent of the 

n u n 2 t 71 1 

angle of the blades at the periphery. 

V — 71 u cot. Wj the pitch of the propeller at the periphery 

360 

X r; 



6 
. 360 

7t uy ^ nuy ^ _ 129600 y ' 



g ' / VQ O ' 



d is the bypothenuse, and y is the base of a right angled plan 
triangle. 

(5 = K C measured over W Y, it is a projection the same as 
U V ^-^ /. 

S^J^^IF^-y^ 

^ 129600 ' 

_ .7854 w^ OB 
' ~~~ 360 

. __ 4 ^ B ((5 + y) 



PRACTICAL MECHANICS. 201 









PROBLEM 1. 

The pitch of the propeller = 33 feet ; and makes .48 
revolutions a minute, the slip ^rzr .35. What is the speed 
of the vessel ? 

^^X^Nl -.35)^11.7 knot. 



88 



By the introduction of the anti-slip screw, the slip would 
be reduced from 35 to 15 per cent., the speed will be 

33 X 48 ,, _ ,. . , . 







88 ^^- ^-^--- 




Put 


=:= the pitch ; 




P 


:^ the revolutions a minute 




n 


= the shp ; and 




J 


= the knots the hour ; 


then 


generally 






i 


-"s^s^C •)^ 




V 


88 j 




— X (1— «) ' 



^^ "~ OX;?' 

88 

~^ _p ( 1 — n) 



202 PRACTICAL MECHANICS. 

A ship fitted up with Byrne and Elliott's Screw made 
13 knots an hour ; the revolutions were 60 a minute, and 
the pitch 30 feet, what was the slip ? 

, 88 i _ 88 X ^^ = .12, or 12 per cent. 

« = ^ — Vxy "" 30 X 60 

PROBLEM 2. 

Given the diameter of a propeller = 12 feet, = w ; and 
the angle at the periphery = 51° ..33' = «;; to find the 

pitch = v1 

Cot.bl° ..33' = w; .7940121 

v^nu cot. w ^ 3.1416 X 12 X .194 = 29.93 feet. 
\ PROBLEM 3. 

Given the diameter = 21 feet = u ; the length of the 
blade parallel with the centre line = 6 feet = r ; the slip 
40 per cent., or s = .40 ; the angle of the blades at the 
periphery = 63° .. 42' = -w ; to find the horse power neces- 
sary to drive this screw 68 revolutions a minute == ex. 

Cosine 63°.. 42' r^ .4430112 
11" X 68- (6 X .4 X .443 + i) = 7124. 



480000 ^ ' 

Log. 21 -- 

log. 68 = 


-- 1.32221 
r 1.832509 




3.154128 
3 


log. 480000 = 


9.464184 
^ 5.681241 


log.oi 1.1124 = 


3.182943 
= .069119 


Log. of 7 1 24 = 


== 3.852122 



PRACTICAL MECHANICS. 203 

As a second example suppose u z= 12 feet ; y = 5 feet ; w = 
bl"" .. 30' ; 5 = 38 per cent., or 5 = 38 ; what power is required 
to drive this propeller 40 revolutions a minute ? 

^^ ' ^ ^^ ' (5 X .38 X .5373 + *) = 509.4 



~ 480000 ''^ '^ 




Log, 15 = 


= 1.176091 


log. 40 = 


- I €02060 




2.178151 




3 



8.334453 
log. 1.13198 = .053839 



8.388292 
log. 480000 = 5.681241 



log. 509.4 = 2.707051 



PROBLEM 4. 

Given the diameter of a propeller = 21 feet = u ; the 
angle of the blades at the periphery = 63^ .. 42' = w ; the 
length y = 6 feet ; the slip s = .40 or 40 per cent. ; four 
engines drive this screw of 7124 horse power = 2;; 
how many revolutions, oc, will this propeller make in a 
minute ? 



1 I *ouuuO z ) ^ ^^ - . 

oc = — ■! -7-. — , ^^ V •" = 68 revolutions. 



480000 . 
{y s cos. w 



y s cos. 'M?-f-i = 6x.4x .443 + i = 1.1743. 



204 PRACTICAL MECHANICS. 

Log. 480000 = 5.681241 
log. 7124 -^ 8.852t24 



9.533965 
log. 1.1U3^ = .069179 



8)9.464186 



3.154729 
log, 21 = 1.322219 



log. 68. = 1.832510 

PROBLEM. 5. 

Given, the diameter of the propeller = 13 feet =u; 
the angle of the blades at the periphery = 59° ,A& = w ; 
the angle of the blades at the hub ^= IP .. 48' = i^, ; the 
diameter of the hub = 1.25 feet, =■■ twice Q S ; (see figure 
16,) to find the pitch at the centre of effort. 

Cotangent of 59^ .. 46' -^ .582793 
Cotangent of 11° .. 48' = 4.786730 

The pitch of the periphery 

= 13 X .583 X 3.1416 = 23.8 feet — v 
The pitch at the hub 

1= 1.25 X 4.787 X 3.1416 = 18.8 feet, v^ 

Put v^ for the pitch at the centre of effort, u^ the diameter at 
the centre of effort, and u^ the diameter at the hub. 

{y — Vg) : (v, — %) : : (^'' — ^^2) • (.7257^ — ^^2) 

. „ -„ I (^ — ^2) (.725 2^ — -z/^J 

• • V, '^2 » 

' '^ u — U^ 



PRACTICAL MECHANICS. 

RECAPITULATION. 



205 



V = 28 . 8 
22 . 3 
18 . 8 



Vj = 22 . 3 



u :^:^ 13 feet 

u^ — ,125 u = 9.425 feet 



^o 



1.25 feet. 



PROBLEM 6. 

Giyen, the diameter of a propeller =: 13.25 feet == u; 
and the angle w = 65°, to find the acting area of the pro- 
peller ? 

5 to' 



but V' 



^ \/ 1'^ -j- n" w 
n"^ w^ cot.^ w, hence 
5u' 



2 -,, 2 > 



A, 



2 7t 7t cosec, w 



log, 13.5 =^ 1.122216 

log, n z^ .491151 

log. cosec. w =: .042724 



1.662091 



log, 2.5 = .397940 
3 log. 13.5 = 3.366648 



3.764588 
1.662091 



log. 126.6 — 2.102497 



.•• Acthig area of propeller =:i^ 126.6 square feet = A, 



206 PRACTICAL MECHANICS. 

The Anti-slip Screw Propeller, Patented by the Au" 
thor of this work, Mr. Oliver Byrne, in conjunction with 
J. G. Elliott, Esq. 5th October, 1858. 

This propeller may be constructed of any of the known 
forms, and operated by any of the known methods of ^eer- 
ing, but the form found to be the best, is that Avhich has 
the' centre of effort in the centre of the acting area of 
each blade ; and when the water is discharged through the 
backs of the blades, the openings should be opposite, or 
symmetrically arranged to these centres. It has been 
found, that the power necessary to turn a solid metal screw 
and shaft, doing a given work, at a given velocity, needs 
not to be increased when a hollow shaft and screw through 
which water flows, are employed to do the same work, un- 
der the same circumstances. 

Openings of different forms may be made in the blades, 
(see A, A, A, A, Fig. 16,) provided the action of the water 
passing through them, has no tendency to give a rotary 
motion of the screw. 

A valve may be placed in any convenient place in the 
hollow shaft, or pipe, that conducts the water aft the pro- 
peller, to prevent its return in cases of back action. 

Fig. 17, is an end view of a four blade screw propeller. 
A, A, A, A, are openings at the rear of the blades, oppo- 
site the centres of effort in the acting areas. A, A, A, A, 
communicate throngli passages with the hollow shaft, A Q, 
(Fig, 16.) 

Fig. 16, is a longitudinal view of Fig. 17. The pro- 
peller is represented as permanently keyed on the shaft, 
but may be retained by a clutch. 



PRACTICAL MECHANICS. 207 

As the propeller is moved, the advanced half of the blade 
6 X W, cuts into the water undisturbed by the action of 
the screw, and as the water passes e, Fig. 17. the aft 
part of the blade ceases to be as efEcient as the fore part 
e W, on account of the thickness of the metal of which 
tlie propeller is formed ; a supply of water through A, ren- 
ders the aft part of the blade as effective in propelling the 
vessel as the far part. When water is conducted aft the 
propeller through a hollow shaft, or through openings in 
the blades, or tlirough both shaft and blades at the same 
time, the slip of the screw is much diminished by the action 
of the blades against the water so discharged. Several 
contrivances may be employed to conduct the useless or 
superfluous water to the rear of the screw propeller, in im- 
mediate contact with the blades aft, not for the purpose of 
giving a rotary motion to the propeller, but for the pur- 
pose of diminishing what is termed slip. 



THE END. 



Nov, 3 1860. 



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